dataset.json

[
    {
        "id": 0,
        "prompt": "Write a Python function to implement a basic singly linked list.",
        "chosen": "class Node:\n    def __init__(self, data=None):\n        self.data = data\n        self.next = None\nclass LinkedList:\n    def __init__(self):\n        self.head = None\n    def print_list(self):\n        node = self.head\n        while node is not None:\n            print(node.data, end=\" \" )\n            node = node.next\n        print()",
        "rejected": "```python\nclass Node:\n    def __init__(self, data):\n        self.data = data\n        self.next = None\n\n\ndef LinkedList():\n    def __init__(self):\n        self.head = None\n\n    def insert(self, data):\n        new_node = Node(data)\n        if self.head is None:\n            self.head = new_node\n        else:\n            current = self.head\n            while current.next is not None:\n                current = current.next\n            current.next = new_node\n\n    def printList(self):\n        current = self.head\n        while current is not None:\n            print(current.data)\n            current = current.next\n\n\n# Example usage:\nlinked_list = LinkedList()\nlinked_list.insert(1)\nlinked_list.insert(2)\nlinked_list.insert(3)\nlinked_list.printList()\n```\n\n**Explanation:**\n\n* **Node** class: Represents a singly linked list node.\n* **LinkedList** class: Implements a singly linked list data structure.\n* **insert()** method: Inserts a new node into the linked list.\n* **printList()** method: Prints the linked list to the console.\n\n**Example Usage:**\n\n* The example code creates a singly linked list object.\n* It then calls the `insert()` method to insert the numbers 1, 2, and 3 into the linked list.\n* Finally, it calls the `printList()` method to print the linked list to the console.\n\n**Output:**\n\n```\n1\n2\n3\n```\n\n**Features:**\n\n* Insertion at the tail of the list.\n* Singly linked list implementation.\n* Basic insertion functionality.\n\n**Time Complexity:**\n\n* Insertion: O(1)\n* Printing: O(n)\n\n**Space Complexity:** O(n), where n is the length of the linked list."
    },
    {
        "id": 1,
        "prompt": "How would you implement a function to insert a new node at the beginning of a linked list?",
        "chosen": "class Node:\n    def __init__(self, data=None):\n        self.data = data\n        self.next = None\ndef insert_at_beginning(head, new_data):\n    new_node = Node(new_data)\n    new_node.next = head\n    head = new_node\n    return head",
        "rejected": "```python\ndef insert_beginning(head, data):\n    new_node = Node(data)\n    new_node.next = head\n    head = new_node\n    return head\n```\n\n**Explanation:**\n\n* `insert_beginning()` function takes two arguments: `head` (the head of the linked list) and `data` (the data to insert).\n* It creates a new node with the given `data`.\n* It sets the `next` pointer of the new node to the current head of the linked list.\n* It updates the head pointer to point to the new node.\n* Finally, it returns the new head of the linked list.\n\n**Example:**\n\n```python\n# Example linked list\nhead = Node(1)\nhead.next = Node(2)\nhead.next.next = Node(3)\n\n# Insert a new node at the beginning\nnew_head = insert_beginning(head, 0)\n\n# Print the linked list\nwhile new_head:\n    print(new_head.data)\n    new_head = new_head.next\n```\n\n**Output:**\n\n```\n0 1 2 3\n```\n\n**Time Complexity:** O(1)\n**Space Complexity:** O(1)\n\n**Note:**\n\n* This implementation assumes that the linked list is singly linked.\n* `Node` is a custom data structure representing a node in the linked list.\n* The function returns the new head of the linked list after insertion."
    },
    {
        "id": 2,
        "prompt": "Write a function to delete a node with a given key from a linked list.",
        "chosen": "class Node:\n    def __init__(self, data=None):\n        self.data = data\n        self.next = None\ndef delete_node(head, key):\n    if head is None:\n        return head\n    if head.data == key:\n        return head.next\n    curr = head\n    while curr.next:\n        if curr.next.data == key:\n            curr.next = curr.next.next\n            return head\n        curr = curr.next\n    return head",
        "rejected": "```python\ndef delete_node(head, key):\n    current = head\n\n    if head is None:\n        return None\n\n    if head.data == key:\n        return None\n\n    previous = None\n\n    while current and current.data != key:\n        previous = current\n        current = current.next\n\n    if current is None:\n        return head\n\n    previous.next = current.next\n    current = None\n\n    return head\n```\n\n**Explanation:**\n\n* **Initialize variables:**\n    * `current` points to the current node in the list.\n    * `previous` points to the previous node in the list.\n\n\n* **Handle special cases:**\n    * If the list is empty, return `None` to indicate an empty list.\n    * If the first node is to be deleted, update `head` to point to the next node in the list.\n\n\n* **Iterate through the list:**\n    * While `current` node is not `None` and its data is not equal to the `key`, update `previous` and `current` to point to the next node in the list.\n\n\n* **Find the node to delete:**\n    * When `current` node's data matches the `key`, break out of the loop.\n\n\n* **Delete the node:**\n    * `previous.next` points to the node after the node to be deleted.\n    * `current` is assigned `None` to indicate that the node has been deleted.\n\n\n* **Return the updated head:**\n    * The function returns `head` of the updated linked list.\n\n**Example:**\n\n```python\n# Example linked list\nhead = Node(1)\nhead.next = Node(2)\nhead.next.next = Node(3)\n\n# Delete the node with key 2\nhead = delete_node(head, 2)\n\n# Print the updated linked list\nwhile head:\n    print(head.data)\n    head = head.next\n```\n\n**Output:**\n\n```\n1\n3\n```\n\n**Time Complexity:** O(n), where n is the length of the linked list.\n\n**Space Complexity:** O(1)"
    },
    {
        "id": 3,
        "prompt": "Implement a function to find the middle element of a linked list.",
        "chosen": "class Node:\n    def __init__(self, data=None):\n        self.data = data\n        self.next = None\ndef find_middle(head):\n    slow = head\n    fast = head\n    while fast and fast.next:\n        slow = slow.next\n        fast = fast.next.next\n    return slow.data",
        "rejected": "```python\ndef middle_element(head):\n    if head is None or head.next is None:\n        return head\n\n    slow = head\n    fast = head\n\n    while fast and fast.next:\n        slow = slow.next\n        fast = fast.next.next\n\n    return slow\n```\n\n**Explanation:**\n\n* **Initialization:**\n    - If the list is empty or has only one element, the middle element is simply the head.\n* **Finding the middle point:**\n    - We use two pointers, `slow` and `fast`, initially set to the head of the list.\n    - We traverse the list until `fast` reaches the end of the list.\n    - During traversal, `slow` advances one node at a time, while `fast` advances two nodes at a time.\n* **When `fast` reaches the end:**\n    - When `fast` reaches the end of the list, `slow` will be pointing to the middle element of the list.\n\n**Example:**\n\n```python\n# Example linked list\nhead = ListNode(1)\nhead.next = ListNode(2)\nhead.next.next = ListNode(3)\nhead.next.next.next = ListNode(4)\nhead.next.next.next.next = ListNode(5)\n\n# Find the middle element\nmiddle_element = middle_element(head)\n\n# Print the middle element\nprint(middle_element.val)  # Output: 3\n```\n\n**Time Complexity:** O(n), where n is the length of the linked list.\n\n**Space Complexity:** O(1)\n\n**Note:**\n\n* This solution is efficient in both time and space complexity.\n* The middle element is identified by traversing the list at different speeds.\n* The function returns the middle element of the linked list."
    },
    {
        "id": 4,
        "prompt": "How would you implement a function to reverse a linked list?",
        "chosen": "class Node:\n    def __init__(self, data=None):\n        self.data = data\n        self.next = None\ndef reverse_list(head):\n    prev = None\n    curr = head\n    while curr:\n        next_node = curr.next\n        curr.next = prev\n        prev = curr\n        curr = next_node\n    return prev",
        "rejected": "**Algorithm to Reverse a Linked List:**\n\n```python\ndef reverseList(head):\n    if head is None or head.next is None:\n        return head\n\n    prev = None\n    curr = head\n\n    while curr:\n        next_node = curr.next\n        curr.next = prev\n        prev = curr\n        curr = next_node\n\n    return prev\n```\n\n**Implementation:**\n\n1. **Initialize two pointers:** `prev` points to the previous node, and `curr` points to the current node in the list.\n2. **Iterate through the list:** While `curr` is not `None`, we:\n    - Store the next node in `next_node`.\n    - Point `curr.next` to `prev`.\n    - Update `prev` to point to `curr`.\n    - Update `curr` to point to `next_node`.\n3. **Return the new head:** The new head of the reversed list is `prev`.\n\n**Example:**\n\n```python\n# Example linked list\nhead = ListNode(1)\nhead.next = ListNode(2)\nhead.next.next = ListNode(3)\nhead.next.next.next = ListNode(4)\n\n# Reverse the linked list\nreversedList = reverseList(head)\n\n# Print the reversed list\nwhile reversedList:\n    print(reversedList.val)\n    reversedList = reversedList.next\n```\n\n**Output:**\n\n```\n4\n3\n2\n1\n```\n\n**Time Complexity:** O(n), where n is the length of the list.\n\n**Space Complexity:** O(1)\n\n**Additional Notes:**\n\n- This algorithm reverses the list in place, without creating a new list.\n- The `prev` pointer points to the previous node in the list, and the `curr` pointer points to the current node in the list.\n- The loop iterates through the list until the `curr` pointer reaches the end of the list.\n- The `prev` pointer is updated to point to the previous node in the list, and the `curr` pointer is updated to point to the next node in the list.\n- Finally, the `prev` node becomes the head of the reversed list."
    },
    {
        "id": 5,
        "prompt": "Write a function to detect a cycle in a linked list.",
        "chosen": "class Node:\n    def __init__(self, data=None):\n        self.data = data\n        self.next = None\ndef has_cycle(head):\n    slow = head\n    fast = head\n    while fast and fast.next:\n        slow = slow.next\n        fast = fast.next.next\n        if slow == fast:\n            return True\n    return False",
        "rejected": "```python\ndef detect_cycle(head):\n    if not head or not head.next:\n        return None\n\n    slow = head\n    fast = head\n\n    while slow and fast and fast.next:\n        slow = slow.next\n        fast = fast.next.next\n\n        if slow == fast:\n            break\n\n    if slow != fast:\n        return None\n\n    slow = head\n\n    while slow != fast:\n        slow = slow.next\n        fast = fast.next\n\n    return slow\n```\n\n**How the function works:**\n\n1. **Initialize two pointers:**\n   - `slow` points to the head of the list.\n   - `fast` points to the head of the list.\n\n\n2. **Iterate through the list until the pointers meet:**\n   - While both `slow` and `fast` are not `None` and `fast`'s next node is not `None`, we move `slow` one step at a time and `fast` two steps at a time.\n   - If `slow` and `fast` ever meet, a cycle is detected.\n\n\n3. **Determine if a cycle exists:**\n   - If `slow` and `fast` do not meet, there is no cycle.\n\n\n4. **Find the starting point of the cycle:**\n   - We reset `slow` to the head of the list.\n   - While `slow` and `fast` are not equal, we move `slow` one step at a time and `fast` one step at a time.\n   - When they finally meet again, the point where they meet is the starting point of the cycle.\n\n\n5. **Return the starting point of the cycle:**\n   - We return the node where the cycle starts.\n\n**Time Complexity:** O(n), where n is the length of the list.\n**Space Complexity:** O(1)\n\n**Example:**\n\n```python\n# Example linked list with a cycle\nhead = ListNode(1)\nhead.next = ListNode(2)\nhead.next.next = ListNode(3)\nhead.next.next.next = head.next\n\n# Detect cycle in the linked list\ncycle_start = detect_cycle(head)\n\n# Print the cycle start point\nprint(cycle_start.val)  # Output: 2\n```"
    },
    {
        "id": 6,
        "prompt": "Implement a basic stack using Python lists. Provide a method to push an element, pop an element, and check if the stack is empty.",
        "chosen": "class Stack:\n    def __init__(self):\n        self.stack = []\n    def push(self, element):\n        self.stack.append(element)\n    def pop(self):\n        if not self.is_empty():\n            return self.stack.pop()\n        else:\n            return None\n    def is_empty(self):\n        return len(self.stack) == 0",
        "rejected": "```python\nclass Stack:\n    def __init__(self):\n        self.stack = []\n\n    def is_empty(self):\n        return len(self.stack) == 0\n\n    def push(self, element):\n        self.stack.append(element)\n\n    def pop(self):\n        if self.is_empty():\n            print(\"Stack is empty\")\n        else:\n            return self.stack.pop()\n```\n\n**Example Usage:**\n\n```python\n# Create a Stack object\nstack = Stack()\n\n# Push elements onto the stack\nstack.push(1)\nstack.push(2)\nstack.push(3)\n\n# Check if the stack is empty\nprint(stack.is_empty())  # Output: False\n\n# Pop elements off the stack\nwhile not stack.is_empty():\n    element = stack.pop()\n    print(element)  # Output: 3 2 1\n\n# Check if the stack is empty again\nprint(stack.is_empty())  # Output: True\n```\n\n**Explanation:**\n\n* **__init__** method initializes an empty list `stack` to represent the empty stack.\n* **is_empty** method checks the length of the `stack` list and returns `True` if the stack is empty or `False` otherwise.\n* **push** method adds an element to the end of the `stack` list.\n* **pop** method removes and returns the top element from the `stack` list.\n\n**Time Complexity:**\n\n* Push: O(1)\n* Pop: O(1)\n* Is Empty: O(1)\n\n**Space Complexity:** O(n), where n is the size of the stack.\n\n**Note:** This implementation uses Python lists as a data structure to implement the stack. Other data structures, such as arrays or linked lists, can also be used for implementing stacks."
    },
    {
        "id": 7,
        "prompt": "Write a function to reverse a stack using a temporary stack in Python.",
        "chosen": "def reverse_stack(stack):\n    temp_stack = []\n    while not stack.is_empty():\n        temp_stack.append(stack.pop())\n    while len(temp_stack) > 0:\n        stack.push(temp_stack.pop())",
        "rejected": "```python\ndef reverseStack(stack):\n    tempStack = []\n\n    while stack:\n        element = stack.pop()\n        while tempStack and element > tempStack[-1]:\n            tempStack.pop()\n        tempStack.append(element)\n\n    while tempStack:\n        stack.append(tempStack.pop())\n\n    return stack\n```\n\n**Explanation:**\n\n1. **Initialize two stacks:** `stack` (input stack) and `tempStack` (temporary stack).\n\n\n2. **Iterate over the input stack:**\n   - While there are elements in `stack`, pop the top element and store it in `element`.\n\n\n3. **Compare elements:**\n   - While `tempStack` is not empty and the `element` is greater than the top element of `tempStack`, pop the top element from `tempStack` and push it back onto `tempStack`. This step ensures that the larger elements remain in the `tempStack`.\n\n\n4. **Push element onto tempStack:**\n   - Finally, push the `element` onto `tempStack`.\n\n\n5. **Transfer elements back to original stack:**\n   - Once the loop completes, while `tempStack` is not empty, pop the top element from `tempStack` and push it onto the `stack` to restore the original order.\n\n\n6. **Return the reversed stack:**\n   - Finally, return the `stack`, which now represents the reversed stack.\n\n**Example:**\n\n```python\n# Example stack\nstack = [1, 2, 3, 4, 5]\n\n# Reverse the stack using the tempStack approach\nreversedStack = reverseStack(stack)\n\n# Print the reversed stack\nprint(reversedStack)  # Output: [5, 4, 3, 2, 1]\n```\n\n**Time Complexity:** O(n), where n is the length of the stack.\n\n**Space Complexity:** O(n), due to the creation of the temporary stack."
    },
    {
        "id": 8,
        "prompt": "Implement a queue using Python lists. Provide a method to enqueue an element, dequeue an element, and check if the queue is empty.",
        "chosen": "class Queue:\n    def __init__(self):\n        self.queue = []\n    def enqueue(self, element):\n        self.queue.append(element)\n    def dequeue(self):\n        if not self.is_empty():\n            return self.queue.pop(0)\n        else:\n            return None\n    def is_empty(self):\n        return len(self.queue) == 0",
        "rejected": "```python\nclass Queue:\n    def __init__(self):\n        self.queue = []\n\n    def enqueue(self, element):\n        self.queue.append(element)\n\n    def dequeue(self):\n        if not self.queue:\n            return None\n        else:\n            return self.queue.pop(0)\n\n    def is_empty(self):\n        return len(self.queue) == 0\n```\n\n**Usage:**\n\n```python\n# Create a queue object\nqueue = Queue()\n\n# Enqueue elements\nqueue.enqueue(1)\nqueue.enqueue(2)\nqueue.enqueue(3)\n\n# Check if the queue is empty\nprint(queue.is_empty())  # Output: False\n\n# Dequeue elements\ndequeued_element = queue.dequeue()\nprint(dequeued_element)  # Output: 1\n\n# Check if the queue is empty again\nprint(queue.is_empty())  # Output: True\n```\n\n**Explanation:**\n\n* **Enqueue()** method appends the given element to the end of the list.\n* **Dequeue()** method removes and returns the first element from the list.\n* **is_empty()** method checks if the list is empty and returns `True` or `False` accordingly.\n\n**Time Complexity:**\n\n* Enqueue: O(1)\n* Dequeue: O(1)\n* Is empty: O(1)\n\n**Space Complexity:** O(n), where n is the length of the queue.\n\n**Note:**\n\n* This implementation is a simple representation of a queue using Python lists. It is suitable for small-scale applications.\n* For larger queues or more complex operations, consider using data structures such as queues or deques from the Python standard library."
    },
    {
        "id": 9,
        "prompt": "Write a function to check if a given string is a palindrome using a stack in Python.",
        "chosen": "def is_palindrome(s):\n    stack = Stack()\n    for char in s:\n        stack.push(char)\n    reversed_s = \"\"\n    while not stack.is_empty():\n        reversed_s += stack.pop()\n    return s == reversed_s",
        "rejected": "```python\ndef isPalindrome(str):\n    stack = []\n    for char in str:\n        stack.append(char)\n\n    while len(stack) > 0:\n        char = stack.pop()\n        if char != str[-1]:\n            return False\n\n    return True\n```\n\n**Explanation:**\n\n* **Initialization:**\n    * A stack `stack` is initialized to store the characters of the string.\n\n\n* **Traversing the string:**\n    * The function iterates over each character in the given string.\n    * The character is pushed onto the stack.\n\n\n* **Checking for palindrome:**\n    * While the stack is not empty:\n        * The top character from the stack is removed and stored in `char`.\n        * It is then compared with the last character of the string.\n        * If the characters do not match, the function returns `False` to indicate that the string is not a palindrome.\n\n\n* **Validation and return:**\n    * If the loop completes without encountering any mismatched characters, it means that the string is a palindrome.\n    * The function returns `True` in this case.\n\n**Example:**\n\n```python\n>>> isPalindrome('racecar')\nTrue\n>>> isPalindrome('hello')\nFalse\n```\n\n**Time Complexity:** O(n), where n is the length of the string.\n\n**Space Complexity:** O(n), due to the use of the stack.\n\n**Additional Notes:**\n\n* This solution uses the Last-In-First-Out (LIFO) property of a stack to efficiently check for palindromes.\n* The function is case-sensitive, so 'racecar' and 'Racecar' are considered different.\n* This implementation does not handle special characters or spaces."
    },
    {
        "id": 10,
        "prompt": "Implement a function to evaluate the validity of a sequence of parentheses in Python.",
        "chosen": "def is_valid_parentheses(s):\n    stack = Stack()\n    for char in s:\n        if char == '(':\n            stack.push(char)\n        elif char == ')':\n            if stack.is_empty():\n                return False\n            stack.pop()\n    return stack.is_empty()",
        "rejected": "```python\ndef is_valid(s):\n    stack = []\n\n    for char in s:\n        if char == '(' or char == '{' or char == '[':\n            stack.append(char)\n        else:\n            if not stack:\n                return False\n\n            top = stack.pop()\n\n            if (char == ')' and top == '(') or (char == '}' and top == '{') or (char == ']' and top == '['):\n                continue\n            else:\n                return False\n\n    return len(stack) == 0\n```\n\n**How it works:**\n\n* The function takes a string `s` representing the sequence of parentheses.\n* It initializes an empty list `stack` to keep track of the encountered parentheses.\n* It iterates over each character in the string:\n    - If the current character is an opening parenthesis (`('`, `{`, or `[`), it is pushed onto the stack.\n    - If the current character is a closing parenthesis (`)', `,` or `]`), it checks if the stack is empty. If it is, the function returns `False` because there is a closing parenthesis without a matching opening parenthesis.\n    - Otherwise, it compares the current closing parenthesis with the top element of the stack. If they match, it removes the top element from the stack and continues the loop. If they don't match, the function returns `False`.\n* Once the loop completes, it checks if the stack is empty. If the stack is empty, the sequence of parentheses is valid. Otherwise, there are unmatched opening parentheses, and the function returns `False`.\n\n**Example:**\n\n```python\n>>> is_valid('()')\nTrue\n>>> is_valid('(()())')\nTrue\n>>> is_valid('(()())()')\nTrue\n>>> is_valid('()())')\nFalse\n>>> is_valid('()(){}')\nFalse\n```\n\n**Time Complexity:** O(n), where n is the length of the string.\n\n**Space Complexity:** O(n), because the function creates a new list `stack` to store the encountered parentheses."
    },
    {
        "id": 11,
        "prompt": "Write a function to convert an infix expression to a postfix expression using a stack in Python.",
        "chosen": "def infix_to_postfix(infix):\n    precedence = {'+': 1, '-': 1, '*': 2, '/': 2}\n    stack = Stack()\n    postfix = \"\"\n    for char in infix:\n        if char.isalpha():\n            postfix += char\n        elif char == '(':\n            stack.push(char)\n        elif char == ')':\n            while not stack.is_empty() and stack.peek() != '(':\n                postfix += stack.pop()\n            stack.pop()\n        else:\n            while not stack.is_empty() and stack.peek() != '(' and precedence[char] <= precedence[stack.peek()]:\n                postfix += stack.pop()\n            stack.push(char)\n    while not stack.is_empty():\n        postfix += stack.pop()\n    return postfix",
        "rejected": "```python\ndef infix_to_postfix(infix_expression):\n    operator_stack = []\n    postfix_expression = []\n\n    for token in infix_expression.split():\n        if token in ['+', '-', '*', '/']:\n            operator_stack.append(token)\n        else:\n            postfix_expression.append(token)\n\n    while operator_stack:\n        operator = operator_stack.pop()\n        postfix_expression.append(operator)\n\n    return ' '.join(postfix_expression)\n```\n\n**Example:**\n\n```python\n>>> infix_expression = \"a + b * c - d / e\"\n>>> infix_to_postfix(infix_expression)\n'a b c * + d e / -'\n```\n\n**How it works:**\n\n1. **Tokenize the infix expression:** The function splits the infix expression into individual tokens (operators and operands).\n\n\n2. **Push operators onto the stack:** While iterating over the tokens, if the token is an operator, it is pushed onto the operator stack.\n\n\n3. **Append operands to the postfix expression:** If the token is an operand, it is appended to the postfix expression.\n\n\n4. **Pop operators from the stack and append to the postfix expression:** Once the iteration is complete, the function pops all remaining operators from the stack and appends them to the postfix expression.\n\n\n5. **Return the postfix expression:** Finally, the function returns the postfix expression.\n\n**Time Complexity:** O(n), where n is the length of the infix expression.\n\n**Space Complexity:** O(n), due to the use of the stack.\n\n**Additional Notes:**\n\n* This function handles only basic arithmetic operators (+, -, *, /).\n* It does not handle parentheses or other special characters.\n* The function returns a space-separated string of the postfix expression."
    },
    {
        "id": 12,
        "prompt": "Write a Python function to perform an in-order traversal of a binary tree.",
        "chosen": "def in_order_traversal(root):\n    if root is not None:\n        in_order_traversal(root.left)\n        print(root.val)\n        in_order_traversal(root.right)",
        "rejected": "```python\ndef inorderTraversal(root):\n    if root is None:\n        return\n\n    inorderTraversal(root.left)\n    print(root.data)\n    inorderTraversal(root.right)\n```\n\n**Explanation:**\n\n* The function takes the root node of the binary tree as input.\n* It first checks if the root node is **None**, if so, it returns.\n* Otherwise, it recursively performs an **inorder traversal** on the **left subtree**.\n* Then, it prints the **data** of the root node.\n* Finally, it recursively performs an inorder traversal on the **right subtree**.\n\n**Example:**\n\n```python\n# Example binary tree\nroot = Node(1)\nroot.left = Node(2)\nroot.right = Node(3)\nroot.left.left = Node(4)\nroot.left.right = Node(5)\n\n# Inorder traversal of the binary tree\ninorderTraversal(root)\n```\n\n**Output:**\n\n```\n4 2 5 1 3\n```\n\n**Time Complexity:** O(n), where n is the number of nodes in the binary tree.\n\n**Space Complexity:** O(h), where h is the height of the binary tree.\n\n**Note:**\n\n* This function performs an **inorder traversal**, which visits the nodes of the binary tree in the following order: left subtree, root, right subtree.\n* The function uses recursion to traverse the binary tree."
    },
    {
        "id": 13,
        "prompt": "How would you implement a post-order traversal of a binary tree in Python?",
        "chosen": "def post_order_traversal(root):\n    if root is not None:\n        post_order_traversal(root.left)\n        post_order_traversal(root.right)\n        print(root.val)",
        "rejected": "```python\ndef postorder(root):\n    if root is None:\n        return\n\n    postorder(root.left)\n    postorder(root.right)\n    print(root.val)\n```\n\n**Explanation:**\n\n* The function takes the root node of the binary tree as an argument.\n* It checks if the root node is None. If so, it returns.\n* Otherwise, it recursively performs the postorder traversal on the left subtree of the root node.\n* Then, it recursively performs the postorder traversal on the right subtree of the root node.\n* Finally, it prints the value of the root node.\n\n**Example:**\n\n```python\n# Example binary tree\nroot = Node(1)\nroot.left = Node(2)\nroot.right = Node(3)\nroot.left.left = Node(4)\nroot.left.right = Node(5)\n\n# Postorder traversal\npostorder(root)\n```\n\n**Output:**\n\n```\n4 5 2 3 1\n```\n\n**Time Complexity:** O(n), where n is the number of nodes in the binary tree.\n\n**Space Complexity:** O(h), where h is the height of the binary tree (depth of the deepest node).\n\n**Note:**\n\n* This implementation uses recursion, which can be inefficient for large binary trees. For larger trees, it is recommended to use an iterative approach using a stack.\n* The `Node` class is a simple representation of a binary tree node with a value and two children."
    },
    {
        "id": 14,
        "prompt": "Write a Python function to perform a pre-order traversal of a binary tree.",
        "chosen": "def pre_order_traversal(root):\n    if root is not None:\n        print(root.val)\n        pre_order_traversal(root.left)\n        pre_order_traversal(root.right)",
        "rejected": "```python\ndef preorder(root):\n    if root is None:\n        return\n\n    print(root.data)\n    preorder(root.left)\n    preorder(root.right)\n```\n\n**Explanation:**\n\n* The function takes the root node of the binary tree as an argument.\n* If the root node is `None`, it returns to indicate that the tree is empty.\n* Otherwise, it prints the data of the root node.\n* Then, it recursively calls itself on the left subtree of the root node.\n* Finally, it recursively calls itself on the right subtree of the root node.\n\n**Example:**\n\n```python\n# Example binary tree\nroot = Node(1)\nroot.left = Node(2)\nroot.right = Node(3)\nroot.left.left = Node(4)\nroot.left.right = Node(5)\n\n# Perform pre-order traversal\npreorder(root)\n```\n\n**Output:**\n\n```\n1\n2\n4\n5\n3\n```\n\n**Time Complexity:** O(n), where n is the number of nodes in the binary tree.\n\n**Space Complexity:** O(h), where h is the height of the binary tree (depth of the deepest node).\n\n**Note:**\n\n* This function performs a pre-order traversal, which visits the nodes in the order root, left subtree, right subtree.\n* The recursive calls to the `preorder` function allow the function to traverse the tree depth-first."
    },
    {
        "id": 15,
        "prompt": "Implement a level-order traversal of a binary tree using a queue in Python.",
        "chosen": "from collections import deque\ndef level_order_traversal(root):\n    if root is None:\n        return\n    queue = deque([root])\n    while queue:\n        node = queue.popleft()\n        print(node.val)\n        if node.left:\n            queue.append(node.left)\n        if node.right:\n            queue.append(node.right)",
        "rejected": "```python\ndef levelOrderTraversal(root):\n    if root is None:\n        return\n\n    queue = [root]\n    while queue:\n        node = queue.pop(0)\n        print(node.val)\n\n        if node.left:\n            queue.append(node.left)\n        if node.right:\n            queue.append(node.right)\n```\n\n**Explanation:**\n\n* **Initialization:**\n    * If the root node is `None`, we return to indicate that the tree is empty.\n\n\n* **Queue-based traversal:**\n    * We initialize a queue and enqueue the root node.\n    * While the queue is not empty, we dequeue the front element and print its value.\n\n\n* **Traversing children:**\n    * If the current node has a left child, we enqueue it to the queue.\n    * If the current node has a right child, we enqueue it to the queue.\n\n\n* **Level-order traversal:**\n    * The order in which we enqueue nodes ensures that we visit them level by level.\n    * The elements at the front of the queue are the nodes at the current level.\n\n\n**Example:**\n\n```python\n# Example binary tree\nroot = Node(1)\nroot.left = Node(2)\nroot.right = Node(3)\nroot.left.left = Node(4)\nroot.left.right = Node(5)\n\n# Level-order traversal\nlevelOrderTraversal(root)\n```\n\n**Output:**\n\n```\n1\n2 3\n4 5\n```\n\n**Time Complexity:** O(n), where n is the number of nodes in the binary tree.\n\n**Space Complexity:** O(n), due to the use of a queue to store the nodes.\n\n**Note:**\n\n* This implementation uses a queue from the `queue` module in Python.\n* The `Node` class is a simple representation of a binary tree node with a value and two children."
    },
    {
        "id": 16,
        "prompt": "How would you implement an iterative in-order traversal of a binary tree in Python?",
        "chosen": "def iterative_in_order_traversal(root):\n    stack = []\n    current = root\n    while True:\n        if current is not None:\n            stack.append(current)\n            current = current.left\n        elif stack:\n            node = stack.pop()\n            print(node.val)\n            current = node.right\n        else:\n            break",
        "rejected": "```python\ndef inorderTraversal(root):\n    if root is None:\n        return\n\n    inorderTraversal(root.left)\n    print(root.val)\n    inorderTraversal(root.right)\n```\n\n**Algorithm:**\n\n1. **Check if the root node is None:** If the root node is empty, return.\n\n\n2. **Inorder traversal of the left subtree:** Recursively traverse the left subtree of the root node.\n\n\n3. **Print the root node value:** Print the value of the root node.\n\n\n4. **Inorder traversal of the right subtree:** Recursively traverse the right subtree of the root node.\n\n**Example:**\n\n```python\n     1\n    / \\\n   2   3\n  / \\   \\\n 4   5   6\n\ninorderTraversal(root)\n```\n\n**Output:**\n\n```\n4\n2\n5\n1\n3\n6\n```\n\n**Time Complexity:** O(n), where n is the number of nodes in the binary tree.\n\n**Space Complexity:** O(h), where h is the height of the binary tree (depth of the deepest node).\n\n**Note:**\n\n* This algorithm performs an iterative in-order traversal of the binary tree.\n* The iterative approach is more efficient than the recursive approach if the height of the binary tree is large.\n* In the case of a complete binary tree of height h, the iterative approach is O(1) in space complexity."
    },
    {
        "id": 17,
        "prompt": "Write a Python function to perform a breadth-first traversal of a binary tree using recursion.",
        "chosen": "def breadth_first_traversal(root):\n    if root is None:\n        return\n    result = []\n    def helper(node, level):\n        if len(result) < level + 1:\n            result.extend([[] for _ in range(level - len(result) + 1)])\n        result[level].append(node.val)\n        if node.left:\n            helper(node.left, level + 1)\n        if node.right:\n            helper(node.right, level + 1)\n    helper(root, 0)\n    return result",
        "rejected": "```python\ndef bfs(root):\n    if root is None:\n        return\n\n    queue = [root]\n    visited = set()\n\n    while queue:\n        node = queue.pop(0)\n        visited.add(node)\n\n        if node.left:\n            queue.append(node.left)\n        if node.right:\n            queue.append(node.right)\n\n    # Perform any desired actions on the nodes after traversal\n```\n\n**Arguments:**\n\n* `root`: The root node of the binary tree.\n\n**Return value:**\n\n* None\n\n**Algorithm:**\n\n1. **Initialize:**\n    - Create a queue to store the nodes to be visited.\n    - Create a set to keep track of the visited nodes.\n\n\n2. **Traversing:**\n    - While the queue is not empty:\n        - Dequeue the first node from the queue.\n        - If the node has not been visited, mark it as visited and enqueue its children.\n\n\n3. **Post-traversal actions:**\n    - Once traversal is complete, you can perform any desired actions on the nodes, such as printing their values or performing computations.\n\n**Example:**\n\n```python\n# Example binary tree\nroot = Node(1)\nroot.left = Node(2)\nroot.right = Node(3)\nroot.left.left = Node(4)\nroot.left.right = Node(5)\n\n# Perform breadth-first traversal\nbfs(root)\n\n# Print the values of the nodes in order\n# Output: [1, 2, 3, 4, 5]\n```\n\n**Time Complexity:** O(n), where n is the number of nodes in the binary tree.\n\n**Space Complexity:** O(n), due to the use of the queue to store the nodes to be visited.\n\n**Note:**\n\n* This function performs a breadth-first traversal using recursion. It is important to note that recursion can be inefficient for large trees. For more efficient breadth-first traversal, it is recommended to use an iterative approach using a queue."
    },
    {
        "id": 18,
        "prompt": "Write a Python function to implement Breadth-First Search (BFS) in a graph.",
        "chosen": "def bfs(graph, root):\n    visited = set()\n    queue = [root]\n    while queue:\n        vertex = queue.pop(0)\n        if vertex not in visited:\n            visited.add(vertex)\n            queue.extend(graph[vertex] - visited)\n    return visited",
        "rejected": "```python\nfrom collections import deque\n\ndef bfs(graph, start):\n    queue = deque([start])\n    visited = set()\n\n    while queue:\n        vertex = queue.popleft()\n        if vertex not in visited:\n            visited.add(vertex)\n            for neighbor in graph[vertex]:\n                queue.append(neighbor)\n\n    return visited\n```\n\n**Arguments:**\n\n* **graph:** A dictionary representing the graph, where keys are vertices and values are lists of neighboring vertices.\n* **start:** The starting vertex for the BFS search.\n\n**Returns:**\n\n* A set containing all the vertices that have been visited during the BFS.\n\n**Example:**\n\n```python\n# Example graph\ngraph = {\n    'A': ['B', 'C'],\n    'B': ['D', 'E'],\n    'C': ['F'],\n    'D': [],\n    'E': [],\n    'F': []\n}\n\n# Perform BFS from vertex 'A'\nvisited = bfs(graph, 'A')\n\n# Print the visited vertices\nprint(visited)\n```\n\n**Output:**\n\n```\n{'A', 'B', 'C', 'D', 'E', 'F'}\n```\n\n**Explanation:**\n\n* The function uses a deque data structure to implement the BFS algorithm.\n* It starts by adding the start vertex to a queue.\n* While the queue is not empty, it removes the first vertex from the queue and marks it as visited.\n* It then iterates over the neighboring vertices of the current vertex and adds them to the queue if they have not been visited.\n* The function returns a set containing all the vertices that have been visited during the BFS.\n\n**Time Complexity:** O(V + E), where V is the number of vertices and E is the number of edges in the graph.\n\n**Space Complexity:** O(V), due to the use of the deque data structure to store the vertices for BFS."
    },
    {
        "id": 19,
        "prompt": "Write a Python function to implement Depth-First Search (DFS) in a graph.",
        "chosen": "def dfs(graph, start):\n    visited = set()\n    traversal_order = []\n    def dfs_helper(vertex):\n        visited.add(vertex)\n        traversal_order.append(vertex)\n        for neighbor in graph[vertex] - visited:\n            dfs_helper(neighbor)\n    dfs_helper(start)\n    return traversal_order",
        "rejected": "```python\nfrom collections import defaultdict\n\ndef dfs(graph, start_node):\n    visited = set()\n    stack = [start_node]\n\n    while stack:\n        node = stack.pop()\n        if node not in visited:\n            visited.add(node)\n            for neighbor in graph[node]:\n                stack.append(neighbor)\n\n    return visited\n```\n\n**Arguments:**\n\n* `graph`: A dictionary representing the graph, where keys are nodes and values are lists of neighboring nodes.\n* `start_node`: The starting node for the DFS.\n\n**Returns:**\n\n* A set containing all the nodes that have been visited during the DFS.\n\n**Example:**\n\n```python\n# Example graph\ngraph = {\n    'A': ['B', 'C'],\n    'B': ['D'],\n    'C': ['E'],\n    'D': [],\n    'E': []\n}\n\n# Perform DFS from node 'A'\nvisited = dfs(graph, 'A')\n\n# Print visited nodes\nprint(visited)\n```\n\n**Output:**\n\n```\n{'A', 'B', 'C', 'D', 'E'}\n```\n\n**Explanation:**\n\n* The function starts by initializing a set to track the visited nodes and a list to represent the nodes to be visited.\n* It iterates over the nodes in the list, checking if they have not been visited.\n* If not visited, it marks the node as visited and recursively calls itself to visit all its neighbors.\n* Finally, the function returns the set of visited nodes.\n\n**Time Complexity:** O(V + E), where V is the number of nodes and E is the number of edges.\n\n**Space Complexity:** O(V), due to the use of the list to keep track of the nodes to be visited."
    },
    {
        "id": 20,
        "prompt": "Write a Python function to implement Dijkstra's algorithm in a graph.",
        "chosen": "import heapq\n\ndef dijkstra(graph, start):\n    distances = {vertex: float('infinity') for vertex in graph}\n    distances[start] = 0\n    priority_queue = [(0, start)]\n    while priority_queue:\n        current_distance, current_vertex = heapq.heappop(priority_queue)\n        for neighbor, weight in graph[current_vertex].items():\n            distance = current_distance + weight\n            if distance < distances[neighbor]:\n                distances[neighbor] = distance\n                heapq.heappush(priority_queue, (distance, neighbor))\n    return distances",
        "rejected": "```python\nimport heapq\n\ndef dijkstra(graph, start, end):\n    distances = {node: float('inf') for node in graph}\n    distances[start] = 0\n\n    queue = [(0, start)]\n    while queue:\n        distance, node = heapq.heappop(queue)\n\n        if node == end:\n            return distances[node]\n\n        for neighbor, weight in graph[node].items():\n            new_distance = distance + weight\n            if new_distance < distances[neighbor]:\n                distances[neighbor] = new_distance\n                heapq.heappush(queue, (new_distance, neighbor))\n\n    return distances[end]\n```\n\n**Arguments:**\n\n* `graph`: A dictionary representing the graph, where the keys are nodes and the values are dictionaries of neighboring nodes and their weights.\n* `start`: The starting node.\n* `end`: The ending node.\n\n**Returns:**\n\n* The shortest path distance between the start and end nodes.\n\n**Example:**\n\n```python\n# Example graph\ngraph = {\n    'A': {'B': 1, 'C': 3},\n    'B': {'C': 2, 'D': 4},\n    'C': {'D': 5, 'E': 6},\n    'D': {'E': 7},\n    'E': {'F': 8},\n}\n\n# Calculate the shortest path distance\ndistance = dijkstra(graph, 'A', 'F')\n\n# Print the result\nprint(distance)  # Output: 15\n```\n\n**Time Complexity:** O(E log V), where E is the number of edges and V is the number of vertices.\n\n**Space Complexity:** O(V), due to the use of a priority queue to store the nodes.\n\n**Note:**\n\n* This implementation uses the heapq library for efficient heap operations.\n* The function returns float('inf') if there is no path between the start and end nodes.\n* This implementation does not handle negative weights."
    },
    {
        "id": 21,
        "prompt": "Write a Python function to implement Topological Sort in a directed acyclic graph (DAG).",
        "chosen": "def topological_sort(graph):\n    in_degree = {vertex: 0 for vertex in graph}\n    for vertex in graph:\n        for neighbor in graph[vertex]:\n            in_degree[neighbor] += 1\n    queue = [vertex for vertex in in_degree if in_degree[vertex] == 0]\n    sorted_order = []\n    while queue:\n        vertex = queue.pop(0)\n        sorted_order.append(vertex)\n        for neighbor in graph[vertex]:\n            in_degree[neighbor] -= 1\n            if in_degree[neighbor] == 0:\n                queue.append(neighbor)\n    return sorted_order",
        "rejected": "```python\nfrom collections import defaultdict\n\ndef topologicalSort(graph):\n    inDegree = defaultdict(int)\n    for u, v in graph.items():\n        inDegree[v] += 1\n\n    queue = [node for node, degree in inDegree.items() if degree == 0]\n    visited = set()\n\n    while queue:\n        node = queue.pop(0)\n        visited.add(node)\n\n        for neighbor in graph[node]:\n            inDegree[neighbor] -= 1\n            if inDegree[neighbor] == 0 and neighbor not in visited:\n                queue.append(neighbor)\n\n    if len(visited) != len(graph):\n        return None\n\n    return list(visited)\n```\n\n**Arguments:**\n\n* `graph`: A dictionary representing the DAG, where keys are nodes and values are lists of neighboring nodes.\n\n**Returns:**\n\n* A list representing the topologically sorted nodes of the DAG, or `None` if the DAG has a cycle.\n\n**Example:**\n\n```python\n# Example DAG\ngraph = {\n    'A': ['B', 'C'],\n    'B': ['D'],\n    'C': ['E'],\n    'D': [],\n    'E': []\n}\n\n# Perform topological sort\nsorted_nodes = topologicalSort(graph)\n\n# Print sorted nodes\nprint(sorted_nodes)  # Output: ['A', 'C', 'B', 'E', 'D']\n```\n\n**Time Complexity:** O(V + E), where V is the number of nodes and E is the number of edges.\n\n**Space Complexity:** O(V), due to the use of `queue` and `visited` sets.\n\n**Note:**\n\n* This function assumes that the input graph is a DAG. If the graph contains cycles, the function will return `None`.\n* The topological sorting algorithm is based on Kahn's algorithm.\n* The `defaultdict` data structure from the `collections` module is used to efficiently maintain the in-degree of nodes."
    },
    {
        "id": 22,
        "prompt": "Write a Python function to implement Bellman-Ford algorithm in a graph.",
        "chosen": "def bellman_ford(graph, source):\n    distance = {vertex: float('inf') for vertex in graph}\n    distance[source] = 0\n    for _ in range(len(graph) - 1):\n        for vertex in graph:\n            for neighbor, weight in graph[vertex].items():\n                distance[neighbor] = min(distance[neighbor], distance[vertex] + weight)\n    for vertex in graph:\n        for neighbor, weight in graph[vertex].items():\n            assert distance[neighbor] <= distance[vertex] + weight, 'Negative cycle detected'\n    return distance",
        "rejected": "```python\nfrom collections import defaultdict\n\ndef bellman_ford(graph, source):\n    distances = defaultdict(float('inf'))\n    distances[source] = 0\n\n    for _ in range(len(graph['edges'])):\n        for u, v, weight in graph['edges']:\n            if distances[u] + weight < distances[v]:\n                distances[v] = distances[u] + weight\n\n    for u, _, weight in graph['edges']:\n        if distances[u] + weight > distances[u]:\n            print(\"Graph contains a negative cycle.\")\n            return\n\n    return distances\n```\n\n**Arguments:**\n\n* **graph**: A dictionary representing the graph, where the keys are nodes and the values are dictionaries with neighboring nodes and edge weights.\n* **source**: The source node to start the algorithm from.\n\n**Returns:**\n\n* A dictionary of distances from the source node to all other nodes in the graph.\n\n**Algorithm:**\n\n1. **Initialization:**\n    - Creates a dictionary to store the distances of each node from the source node.\n    - Sets the distance of the source node to 0.\n\n\n2. **Relaxation Iteration:**\n    - Iterates over the edges in the graph multiple times.\n    - For each edge, it checks if the distance of the source node plus the weight of the edge is less than the distance of the target node.\n    - If true, it updates the distance of the target node.\n\n\n3. **Cycle Detection:**\n    - After the relaxation iterations, it checks for negative cycles by iterating over the edges again.\n    - If it detects a cycle with a negative weight, it prints an error message and returns.\n\n\n4. **Distance Retrieval:**\n    - Finally, it returns the distances of all nodes from the source node.\n\n**Example Usage:**\n\n```python\n# Example graph\ngraph = {\n    'A': {'B': 1, 'C': 3},\n    'B': {'C': 1, 'D': 2},\n    'C': {'D': 4},\n    'D': {'E': 2},\n}\n\n# Perform Bellman-Ford algorithm\ndistances = bellman_ford(graph, 'A')\n\n# Print results\nprint(distances)\n```\n\n**Output:**\n\n```\n{'A': 0, 'B': 1, 'C': 4, 'D': 6, 'E': 8}\n```\n\n**Note:**\n\n* The algorithm handles weighted edges and directed graphs.\n* It detects and handles negative cycles.\n* The complexity of the algorithm is O(E + V), where E is the number of edges and V is the number of vertices."
    },
    {
        "id": 23,
        "prompt": "Write a Python function to implement Floyd-Warshall algorithm in a graph.",
        "chosen": "def floyd_warshall(graph):\n    vertices = list(graph.keys())\n    distance_matrix = [[float('inf') for _ in range(len(vertices))] for _ in range(len(vertices))]\n    for i, vertex in enumerate(vertices):\n        distance_matrix[i][i] = 0\n        for neighbor, weight in graph[vertex].items():\n            distance_matrix[vertices.index(vertex)][vertices.index(neighbor)] = weight\n    for k in range(len(vertices)):\n        for i in range(len(vertices)):\n            for j in range(len(vertices)):\n                distance_matrix[i][j] = min(distance_matrix[i][j], distance_matrix[i][k] + distance_matrix[k][j])\n    return distance_matrix",
        "rejected": "```python\nimport collections\n\ndef floyd_warshall(graph):\n    \"\"\"\n    Implements Floyd-Warshall algorithm in a graph.\n\n    Args:\n        graph: A dictionary representing the graph, where keys are nodes and values are dictionaries of neighboring nodes and distances.\n\n    Returns:\n        A dictionary of distances between nodes.\n    \"\"\"\n\n    # Initialize distances between nodes.\n    distances = collections.deepcopy(graph)\n    for u in graph:\n        for v in graph:\n            distances[(u, v)] = float('inf') if u != v else 0\n\n    # Iterate over all pairs of nodes.\n    for k in range(len(graph)):\n        for u in graph:\n            for v in graph:\n                if u != v and u != k and v != k and distances[(u, k)] + distances[(k, v)] < distances[(u, v)]:\n                    distances[(u, v)] = distances[(u, k)] + distances[(k, v)]\n\n    return distances\n```\n\n**Example Usage:**\n\n```python\n# Example graph.\ngraph = {\n    'A': {'B': 1, 'C': 3},\n    'B': {'C': 2, 'D': 4},\n    'C': {'D': 5},\n    'D': {'C': 1, 'E': 2},\n    'E': {'D': 3},\n}\n\n# Calculate distances using Floyd-Warshall algorithm.\ndistances = floyd_warshall(graph)\n\n# Print distances.\nfor u, v in distances:\n    print(f\"Distance between {u} and {v}: {distances[(u, v)]}\")\n```\n\n**Output:**\n\n```\nDistance between A and A: 0.0\nDistance between A and B: 1.0\nDistance between A and C: 3.0\nDistance between A and D: 6.0\nDistance between A and E: 5.0\nDistance between B and A: 1.0\nDistance between B and B: 0.0\nDistance between B and C: 2.0\nDistance between B and D: 4.0\nDistance between B and E: 3.0\nDistance between C and A: 3.0\nDistance between C and B: 2.0\nDistance between C and C: 1.0\nDistance between C and D: 5.0\nDistance between C and E: 4.0\nDistance between D and A: 6.0\nDistance between D and B: 4.0\nDistance between D and C: 5.0\nDistance between D and D: 0.0\nDistance between D and E: 2.0\nDistance between E and A: 5.0\nDistance between E and B: 3.0\nDistance between E and C: 4.0\nDistance between E and D: 2.0\nDistance between E and E: 0.0\n```\n\n**Time Complexity:** O(V^3), where V is the number of nodes in the graph.\n\n**Space Complexity:** O(V^2), due to the storage of the distances between nodes."
    },
    {
        "id": 24,
        "prompt": "Write a Python function to calculate the nth Fibonacci number using dynamic programming?",
        "chosen": "def fibonacci(n):\n    fib = [0]*(n+1)\n    fib[1] = 1\n    for i in range(2, n+1):\n        fib[i] = fib[i-1] + fib[i-2]\n    return fib[n]",
        "rejected": "```python\ndef fibonacci(n):\n    \"\"\"Calculates the nth Fibonacci number using dynamic programming.\n\n    Args:\n        n (int): The index of the Fibonacci number to calculate.\n\n    Returns:\n        int: The nth Fibonacci number.\n    \"\"\"\n\n    # Initialize a dictionary to store the cached Fibonacci numbers.\n    cache = {0: 0, 1: 1}\n\n    # Check if the Fibonacci number is already cached.\n    if n in cache:\n        return cache[n]\n\n    # Calculate the nth Fibonacci number.\n    fib = fibonacci(n-1) + fibonacci(n-2)\n\n    # Cache the Fibonacci number.\n    cache[n] = fib\n\n    # Return the Fibonacci number.\n    return fib\n```\n\n**Example Usage:**\n\n```python\n# Calculate the 5th Fibonacci number.\nfib = fibonacci(5)\n\n# Print the result.\nprint(fib)  # Output: 5\n```\n\n**How the Function Works:**\n\n1. **Initialization:**\n   - The function initializes a dictionary called `cache` to store the cached Fibonacci numbers.\n   - It sets the base cases: `cache[0]` and `cache[1]` to 0 and 1 respectively.\n\n\n2. **Caching:**\n   - If the Fibonacci number `n` is already cached, the function returns the cached value.\n\n\n3. **Calculation:**\n   - If the Fibonacci number `n` is not cached, the function recursively calculates it by adding the two previous Fibonacci numbers, which are retrieved from the `cache` dictionary.\n\n\n4. **Caching the Result:**\n   - After calculating the Fibonacci number, the function caches it in the `cache` dictionary under the key `n`.\n\n\n5. **Return Value:**\n   - Finally, the function returns the calculated Fibonacci number.\n\n**Time Complexity:** O(1)\n\n**Space Complexity:** O(1)\n\n**Note:**\n\n- The dynamic programming approach to Fibonacci number calculation avoids the need for recursion, which can be inefficient for large values of `n`.\n- The cached Fibonacci numbers are stored in a dictionary, which provides efficient access and avoids redundant calculations."
    },
    {
        "id": 25,
        "prompt": "Implement a dynamic programming solution to find the longest common subsequence of two strings?",
        "chosen": "def longest_common_subsequence(str1, str2):\n    m, n = len(str1), len(str2)\n    dp = [[0]*(n+1) for _ in range(m+1)]\n    for i in range(m+1):\n        for j in range(n+1):\n            if i == 0 or j == 0:\n                dp[i][j] = 0\n            elif str1[i-1] == str2[j-1]:\n                dp[i][j] = dp[i-1][j-1] + 1\n            else:\n                dp[i][j] = max(dp[i-1][j], dp[i][j-1])\n    return dp[m][n]",
        "rejected": "```python\ndef longest_common_subsequence(str1, str2):\n    len1, len2 = len(str1), len(str2)\n    lcs = [[0 for _ in range(len2 + 1)] for _ in range(len1 + 1)]\n\n    for i in range(1, len1 + 1):\n        for j in range(1, len2 + 1):\n            if str1[i - 1] == str2[j - 1]:\n                lcs[i][j] = lcs[i - 1][j - 1] + 1\n            else:\n                lcs[i][j] = max(lcs[i][j - 1], lcs[i - 1][j])\n\n    return lcs[len1][len2]\n```\n\n**How it works:**\n\n* **Initialization:**\n    - Creates a 2D matrix `lcs` of size `len1+1`x`len2+1` to store the length of the longest common subsequence for each pair of characters.\n    - Initialization values are all 0.\n\n\n* **Dynamic Programming:**\n    - Iterates over the rows and columns of the `lcs` matrix.\n    - For each cell, it checks if the last characters of the two strings match.\n        - If they match, it recursively calculates the length of the longest common subsequence of the remaining strings.\n        - If they don't match, it takes the maximum of the lengths of the longest common subsequences of the shorter strings.\n\n\n* **Result:**\n    - The value in the bottom-right corner of the `lcs` matrix represents the length of the longest common subsequence of the two strings.\n\n**Example:**\n\n```python\nstr1 = \"abcde\"\nstr2 = \"ace\"\n\nlen = longest_common_subsequence(str1, str2)\n\nprint(len)  # Output: 3\n```\n\n**Time Complexity:** O(m*n), where m and n are the lengths of the two strings.\n\n**Space Complexity:** O(m*n), due to the creation of the `lcs` matrix."
    },
    {
        "id": 26,
        "prompt": "Write a dynamic programming solution to find the minimum number of coins needed to make change for a given amount?",
        "chosen": "def min_coins(coins, amount):\n    dp = [float('inf')] * (amount + 1)\n    dp[0] = 0\n    for coin in coins:\n        for i in range(coin, amount + 1):\n            dp[i] = min(dp[i], dp[i - coin] + 1)\n    return dp[amount] if dp[amount] != float('inf') else -1",
        "rejected": "```python\ndef coin_change(amount, coins):\n    \"\"\"\n    Dynamic programming solution to find the minimum number of coins needed to make change for a given amount.\n\n    Args:\n        amount (int): The amount of money to make change for.\n        coins (list): A list of coin denominations.\n\n    Returns:\n        int: The minimum number of coins needed to make change.\n    \"\"\"\n\n    # Create a table to store the minimum number of coins needed to make change for each amount.\n    dp = [float('inf') for i in range(amount + 1)]\n    dp[0] = 0\n\n    # Iterate over the amounts from 1 to the given amount.\n    for i in range(1, amount + 1):\n        for j in coins:\n            if i - j >= 0:\n                dp[i] = min(dp[i], dp[i - j] + 1)\n\n    return dp[amount]\n```\n\n**Example:**\n\n```python\namount = 10\ncoins = [1, 2, 5]\n\nmin_coins = coin_change(amount, coins)\n\nprint(min_coins)  # Output: 3\n```\n\n**Time Complexity:** O(n * m), where n is the amount and m is the number of coins.\n\n**Space Complexity:** O(n), due to the creation of the dp table.\n\n**Algorithm:**\n\n1. Create a table (dp) to store the minimum number of coins needed to make change for each amount from 0 to the given amount.\n2. Initialize the table with infinity values for all amounts except 0, which is set to 0.\n3. Iterate over the amounts from 1 to the given amount.\n4. For each amount, iterate over the coins.\n5. If the current amount minus the coin is greater than or equal to 0, update the dp table for the current amount to the minimum of the current value and the value in the dp table for the amount minus the coin plus 1.\n6. Return the value in the dp table for the given amount.\n\n**Note:**\n\n* This solution assumes that the coins are of different denominations.\n* The coins are assumed to be infinite in supply.\n* The amount and coins are both integers."
    },
    {
        "id": 27,
        "prompt": "Implement a dynamic programming solution to find the maximum value that can be put in a knapsack of capacity W?",
        "chosen": "def knapsack(weights, values, capacity):\n    n = len(values)\n    dp = [[0 for _ in range(capacity + 1)] for _ in range(n + 1)]\n    for i in range(n + 1):\n        for w in range(capacity + 1):\n            if i == 0 or w == 0:\n                dp[i][w] = 0\n            elif weights[i-1] <= w:\n                dp[i][w] = max(values[i-1] + dp[i-1][w-weights[i-1]], dp[i-1][w])\n            else:\n                dp[i][w] = dp[i-1][w]\n    return dp[n][capacity]",
        "rejected": "```python\ndef knapsack(values, weights, W):\n    n = len(values)\n    knapsack = [0] * (W + 1)\n\n    for i in range(1, n + 1):\n        for w in range(W, -1, -1):\n            if w >= weights[i - 1]:\n                knapsack[w] = max(knapsack[w], knapsack[w - weights[i - 1]] + values[i - 1])\n\n    return knapsack[W]\n```\n\n**Explanation:**\n\n* **Initialization:**\n    - Create a knapsack array of size W + 1, initialized to 0. This represents the maximum value that can be accommodated in the knapsack of capacity W.\n* **Iterating over items:**\n    - Loop through each item in the list.\n* **Iterating over knapsack capacity:**\n    - Loop through the knapsack capacity in descending order, starting from W.\n* **Checking item weight:**\n    - If the current knapsack capacity is greater than or equal to the item weight, update the knapsack array at that capacity to the maximum of its current value and the value of the item plus the previous knapsack value at capacity W - weight.\n* **Result:**\n    - Return the value at the knapsack capacity W.\n\n**Example:**\n\n```python\nvalues = [60, 10, 12, 13, 18]\nweights = [10, 2, 3, 4, 5]\nW = 50\n\nmax_value = knapsack(values, weights, W)\n\nprint(max_value)  # Output: 135\n```\n\n**Time Complexity:** O(n * w), where n is the number of items and w is the knapsack capacity.\n\n**Space Complexity:** O(w), due to the knapsack array.\n\n**Key Features:**\n\n* Dynamic programming approach.\n* Handles multiple items and knapsack capacity.\n* Returns the maximum value that can be accommodated in the knapsack.\n\n**Applications:**\n\n* Resource allocation.\n* Portfolio optimization.\n* Cutting stock optimization."
    },
    {
        "id": 28,
        "prompt": "Write a dynamic programming solution to find the length of the longest increasing subsequence in a given array?",
        "chosen": "def longest_increasing_subsequence(arr):\n    n = len(arr)\n    dp = [1] * n\n    for i in range(1, n):\n        for j in range(i):\n            if arr[i] > arr[j] and dp[i] < dp[j] + 1:\n                dp[i] = dp[j] + 1\n    return max(dp)",
        "rejected": "```python\ndef lengthOfLIS(nums):\n    if not nums:\n        return 0\n\n    n = len(nums)\n    lis = [1] * n\n\n    for i in range(1, n):\n        for j in range(i):\n            if nums[i] > nums[j] and lis[i] < lis[j] + 1:\n                lis[i] = lis[j] + 1\n\n    return lis[n - 1]\n```\n\n**Algorithm:**\n\n1. **Initialization:**\n   - Create a list `lis` of length `n`, initialized to 1, representing the length of the longest increasing subsequence for each element.\n\n\n2. **Dynamic Programming Iteration:**\n   - Iterate over the elements of the array.\n   - For each element, iterate over the previous elements.\n   - If the current element is greater than the previous element and the length of the longest increasing subsequence for the previous element is greater than the current element's subsequence, update the current element's subsequence length to the previous element's subsequence length plus 1.\n\n\n3. **Result:**\n   - Return the length of the longest increasing subsequence for the last element in the `lis` list.\n\n\n**Example:**\n\n```python\nnums = [10, 9, 2, 5, 3, 7, 101, 18]\nresult = lengthOfLIS(nums)\nprint(result)  # Output: 4\n```\n\n**Time Complexity:** O(n^2), where n is the length of the input list.\n\n**Space Complexity:** O(n), for the `lis` list.\n\n**Key Features:**\n\n- This solution uses dynamic programming to cache the results of previous computations, avoiding redundant calculations.\n- The `lis` list stores the length of the longest increasing subsequence for each element.\n- The algorithm efficiently updates the `lis` list by considering only the previous elements.\n\n**Variations:**\n\n- **Faster Solutions:** For larger input sizes, more efficient solutions exist using binary search or other data structures.\n- **Additional Criteria:** To find the elements in the longest increasing subsequence, additional bookkeeping is required."
    },
    {
        "id": 29,
        "prompt": "Implement a dynamic programming solution to solve the 0/1 knapsack problem with a twist: the capacity of the knapsack increases by a fixed amount for each item added?",
        "chosen": "def dynamic_knapsack(weights, values, capacity, increment):\n    n = len(values)\n    dp = [[0 for _ in range(capacity + 1)] for _ in range(n + 1)]\n    for i in range(n + 1):\n        for w in range(capacity + 1):\n            if i == 0 or w == 0:\n                dp[i][w] = 0\n            elif weights[i-1] <= w:\n                dp[i][w] = max(values[i-1] + dp[i-1][max(0, w-weights[i-1]-increment)], dp[i-1][w])\n            else:\n                dp[i][w] = dp[i-1][w]\n    return dp[n][capacity]",
        "rejected": "```python\ndef knapsack_dynamic(weights, values, capacities, fixed_capacity):\n    n, c = len(weights), len(capacities)\n    dp = [0] * (c + fixed_capacity)\n\n    for i in range(n):\n        for j in range(c + fixed_capacity, -1, -1):\n            if j >= weights[i]:\n                dp[j] = max(dp[j], dp[j - weights[i]] + values[i])\n\n    return dp[c + fixed_capacity]\n```\n\n**Problem Statement:**\n\nGiven a set of items with weights and values, and a knapsack with a variable capacity, find the maximum value that can be achieved by selecting a subset of the items such that the total weight does not exceed the knapsack's capacity.\n\n**Solution:**\n\nThe solution is based on dynamic programming. We define `dp[j]` as the maximum value that can be achieved with a knapsack capacity of `j`.\n\n**Algorithm:**\n\n1. Initialize `dp` table with zeros for all capacities.\n\n2. Iterate over items.\n3. Iterate over knapsack capacities in descending order, starting from the capacity of the knapsack plus the fixed capacity.\n4. For each capacity, if the current item's weight is less than the capacity, we update `dp[j]` to the maximum of its current value and the value of the solution with the current item added.\n\n**Example:**\n\n```python\nweights = [2, 3, 5]\nvalues = [6, 10, 12]\ncapacities = [5, 7]\nfixed_capacity = 2\n\nknapsack_value = knapsack_dynamic(weights, values, capacities, fixed_capacity)\n\nprint(knapsack_value)  # Output: 22\n```\n\n**Time Complexity:** O(n * c), where n is the number of items and c is the knapsack capacity.\n\n**Space Complexity:** O(c), due to the creation of the `dp` table.\n\n**Note:**\n\n* The fixed capacity is added to the knapsack capacity for each item.\n* This solution is suitable for knapsack problems where the knapsack capacity can be increased by a fixed amount for each item."
    },
    {
        "id": 30,
        "prompt": "Implement a Python function to solve the activity selection problem using a greedy algorithm. The function should take a list of activities with their start and end times as input and return the maximum number of activities that can be performed by a single person.",
        "chosen": "def activity_selection(activities):\n    activities.sort(key=lambda x: x[1])\n    selected_activities = [activities[0]]\n    for activity in activities[1:]:\n        if activity[0] >= selected_activities[-1][1]:\n            selected_activities.append(activity)\n    return selected_activities",
        "rejected": "```python\ndef activitySelection(activities):\n    activities.sort(key=lambda x: x[1])  # Sort activities by end time\n\n    selected = []\n    current = 0\n\n    for i in range(len(activities)):\n        if activities[i][0] >= activities[current][1]:\n            selected.append(i)\n            current = i\n\n    return len(selected)\n```\n\n**Explanation:**\n\n* **Sorting:** The activities are sorted based on their end times in ascending order. This ensures that we can always choose the next activity without violating the constraint that it must end before the previous one ends.\n* **Selection Loop:** The function iterates over the list of activities.\n* **Selection Criteria:** If the current activity's start time is greater than or equal to the previous activity's end time, it means that the current activity can be added without violating the constraint.\n* **Adding to Selection:** If the current activity meets the selection criteria, it is added to the `selected` list and the index of the current activity is stored in `current`.\n* **Output:** The function returns the length of the `selected` list, which represents the maximum number of activities that can be performed without violating the constraints.\n\n**Example:**\n\n```python\nactivities = [\n    [2, 3],\n    [1, 2],\n    [3, 4],\n    [6, 7],\n]\n\nresult = activitySelection(activities)\n\nprint(result)  # Output: 3\n```\n\n**Time Complexity:** O(n log n), where n is the length of the list of activities.\n\n**Space Complexity:** O(n), due to the creation of the `selected` list.\n\n**Note:** This solution assumes that the activities are mutually exclusive. If multiple activities can overlap, a different approach is required to solve the problem efficiently."
    },
    {
        "id": 31,
        "prompt": "Write a Python function to solve the Huffman coding problem using a greedy algorithm. The function should take a string as input and return the encoded string and the Huffman tree.",
        "chosen": "import heapq\n\ndef calculate_frequency(string):\n    frequency = {}\n    for symbol in string:\n        if symbol not in frequency:\n            frequency[symbol] = 0\n        frequency[symbol] += 1\n    return frequency\n\ndef build_heap(frequency):\n    heap = []\n    for key in frequency:\n        node = HeapNode(key, frequency[key])\n        heapq.heappush(heap, node)\n    return heap\n\ndef merge_nodes(heap):\n    while len(heap) > 1:\n        node1 = heapq.heappop(heap)\n        node2 = heapq.heappop(heap)\n        merged = HeapNode(None, node1.freq + node2.freq)\n        merged.left = node1\n        merged.right = node2\n        heapq.heappush(heap, merged)\n    return heap\n\ndef build_codes_helper(root, current_code, codes):\n    if root == None:\n        return\n    if root.char != None:\n        codes[root.char] = current_code\n    build_codes_helper(root.left, current_code + \"0\", codes)\n    build_codes_helper(root.right, current_code + \"1\", codes)\n\ndef build_codes(root):\n    codes = {}\n    build_codes_helper(root, \"\", codes)\n    return codes\n\ndef huffman_encoding(string):\n    frequency = calculate_frequency(string)\n    heap = build_heap(frequency)\n    merge_nodes(heap)\n    root = heap[0]\n    codes = build_codes(root)\n    encoded_string = \"\"\n    for char in string:\n        encoded_string += codes[char]\n    return encoded_string, root",
        "rejected": "```python\nimport collections\n\ndef huffman_encode(string):\n    char_counts = collections.Counter(string)\n    characters, frequencies = list(char_counts.items())\n    characters.sort(key=lambda x: x[1], reverse=True)\n\n    # Build the Huffman tree\n    parents = {}\n    tree = {}\n    queue = [(frequency, character) for character, frequency in characters]\n\n    while len(queue) > 1:\n        frequency1, character1 = queue.pop(0)\n        frequency2, character2 = queue.pop(0)\n        new_frequency = frequency1 + frequency2\n        new_character = character1 + character2\n        parents[new_character] = (character1, character2)\n        tree[new_character] = new_frequency\n        queue.append((new_frequency, new_character))\n\n    # Encode the string\n    encoded_string = ''\n    current = ''\n    for char in string:\n        if char not in tree:\n            continue\n        current = parents[char]\n        while current in parents:\n            encoded_string = encoded_string + current[-1]\n            current = parents[current]\n\n    return encoded_string, tree\n```\n\n**Example Usage:**\n\n```python\nstring = 'aaabbccccdddeeeefff'\nencoded_string, tree = huffman_encode(string)\n\nprint('Encoded string:', encoded_string)\nprint('Huffman tree:', tree)\n```\n\n**Output:**\n\n```\nEncoded string: 11110001110011100111\nHuffman tree: {'a': '1', 'b': '0', 'c': '11', 'd': '10', 'e': '101', 'f': '100'}\n```\n\n**Explanation:**\n\n* The function starts by counting the frequency of each character in the input string.\n* It then builds a Huffman tree based on the character frequencies, where the nodes are arranged in a binary tree structure.\n* The encoding process starts by traversing the Huffman tree, starting from the root node.\n* Each character is assigned a binary code based on the path from the root to the leaf node representing that character.\n* The encoded string is constructed by traversing the tree and following the assigned binary codes for each character in the input string."
    },
    {
        "id": 32,
        "prompt": "Implement a Python function to solve the 0/1 knapsack problem using a greedy algorithm. The function should take a list of items with their weights and values as input and return the maximum value that can be put in a knapsack of a given capacity.",
        "chosen": "def knapsack_greedy(items, capacity):\n    items.sort(key=lambda x: x[1] / x[0], reverse=True)\n    total_value = 0\n    for item in items:\n        if item[0] <= capacity:\n            capacity -= item[0]\n            total_value += item[1]\n        else:\n            fractional_weight = capacity / item[0]\n            total_value += item[1] * fractional_weight\n            break\n    return total_value",
        "rejected": "```python\ndef knapsack(items, capacity):\n    items.sort(key=lambda x: x[1]/x[0], reverse=True)\n    total_value = 0\n    total_weight = 0\n\n    for item in items:\n        if total_weight + item[0] <= capacity:\n            total_value += item[1]\n            total_weight += item[0]\n        else:\n            break\n\n    return total_value\n```\n\n**Arguments:**\n\n* `items`: A list of items, where each item is a tuple of (weight, value).\n* `capacity`: The capacity of the knapsack.\n\n**Return Value:**\n\n* The maximum value that can be put in the knapsack.\n\n**Algorithm:**\n\n1. **Sort the items** in descending order of value-to-weight ratio.\n2. **Iterate over the items** and add them to the knapsack until the knapsack capacity is reached.\n3. **Return the total value** of the items added to the knapsack.\n\n**Example:**\n\n```python\nitems = [\n    (2, 10),\n    (3, 20),\n    (4, 30),\n    (5, 40),\n]\ncapacity = 5\n\nknapsack_value = knapsack(items, capacity)\n\nprint(knapsack_value)  # Output: 50\n```\n\n**Time Complexity:** O(n log n), where n is the number of items.\n\n**Space Complexity:** O(1)\n\n**Note:**\n\n* This algorithm is a greedy algorithm, which means it makes the best local decision at each step without considering the future consequences.\n* The sorting step is essential for the efficiency of the algorithm.\n* The knapsack problem is NP-hard, meaning that it is computationally difficult to solve for large instances."
    },
    {
        "id": 33,
        "prompt": "Write a Python function to solve the Coin Changing problem using a greedy algorithm. The function should take a list of coin denominations and a target amount as input and return the minimum number of coins required to make the target amount.",
        "chosen": "def coin_changing_greedy(coins, target):\n    coins.sort(reverse=True)\n    num_coins = 0\n    for coin in coins:\n        while target >= coin:\n            target -= coin\n            num_coins += 1\n    return num_coins",
        "rejected": "```python\ndef coin_change(coins, amount):\n    \"\"\"\n    Solves the Coin Changing problem using a greedy algorithm.\n\n    Args:\n        coins (list): A list of coin denominations.\n        amount (int): The target amount.\n\n    Returns:\n        int: The minimum number of coins required to make the target amount.\n    \"\"\"\n\n    coins.sort(reverse=True)  # Sort coins in descending order of value\n\n    num_coins = 0\n    current_amount = amount\n\n    for coin in coins:\n        while current_amount >= coin:\n            current_amount -= coin\n            num_coins += 1\n\n    return num_coins\n```\n\n**Example Usage:**\n\n```python\n# Example coins and amount\ncoins = [10, 5, 2, 1]\namount = 11\n\n# Calculate coin change\nnum_coins = coin_change(coins, amount)\n\n# Print result\nprint(\"Number of coins required:\", num_coins)\n```\n\n**Output:**\n\n```\nNumber of coins required: 3\n```\n\n**Algorithm:**\n\n1. Sort the coins in descending order of value.\n2. Iterate over the coins in order.\n3. While the current amount is greater than or equal to the current coin, deduct the coin value from the current amount and increment the number of coins used.\n4. Repeat steps 2 and 3 until the current amount reaches 0.\n\n**Time Complexity:** O(n), where n is the length of the coins list.\n\n**Space Complexity:** O(1)\n\n**Note:**\n\n* This solution uses a greedy approach, which does not guarantee the optimal solution for all cases.\n* The coins list should be a list of positive integers representing the denominations of the coins.\n* The amount parameter should be a positive integer representing the target amount."
    },
    {
        "id": 34,
        "prompt": "Implement a Python function to solve the Fractional Knapsack problem using a greedy algorithm. The function should take a list of items with their weights and values as input and return the maximum value that can be put in a knapsack of a given capacity.",
        "chosen": "def fractional_knapsack_greedy(items, capacity):\n    items.sort(key=lambda x: x[1] / x[0], reverse=True)\n    total_value = 0\n    for item in items:\n        if item[0] <= capacity:\n            capacity -= item[0]\n            total_value += item[1]\n        else:\n            fractional_weight = capacity / item[0]\n            total_value += item[1] * fractional_weight\n            break\n    return total_value",
        "rejected": "```python\ndef fractional_knapsack(items, capacity):\n    items.sort(key=lambda x: x[1]/x[0], reverse=True)\n\n    total_value = 0\n    total_weight = 0\n\n    for item in items:\n        if total_weight + (item[0] / item[1]) <= capacity:\n            total_value += item[1]\n            total_weight += item[0] / item[1]\n        else:\n            break\n\n    return total_value\n```\n\n**Arguments:**\n\n* `items`: A list of items, where each item is a tuple containing its weight and value.\n* `capacity`: The capacity of the knapsack.\n\n**Return Value:**\n\n* The maximum value that can be put in the knapsack.\n\n**Algorithm:**\n\n1. **Sort the items** in descending order of value-to-weight ratio.\n2. **Iterate over the items** in the sorted list.\n3. **Check if the item's weight can be partially added** to the knapsack without exceeding the capacity.\n4. **If possible**, add the item to the knapsack and update the total weight and value.\n5. **If not possible**, break out of the loop.\n6. **Return the total value** of the items added to the knapsack.\n\n**Example:**\n\n```python\nitems = [\n    (2, 10),\n    (3, 20),\n    (4, 30),\n]\ncapacity = 5\n\nmax_value = fractional_knapsack(items, capacity)\n\nprint(max_value)  # Output: 29.33\n```\n\n**Time Complexity:** O(n log n), where n is the number of items.\n\n**Space Complexity:** O(1)\n\n**Note:**\n\n* This solution is a greedy algorithm and may not always find the optimal solution for the Fractional Knapsack problem.\n* The value-to-weight ratio is used as a proxy for the item's importance."
    },
    {
        "id": 35,
        "prompt": "Write a Python function to solve the Scheduling Problem using a greedy algorithm. The function should take a list of tasks with their deadlines as input and return the maximum number of tasks that can be completed.",
        "chosen": "def scheduling_greedy(tasks):\n    tasks.sort(key=lambda x: x[1])\n    completed_tasks = 0\n    current_time = 0\n    for task in tasks:\n        if current_time + task[0] <= task[1]:\n            current_time += task[0]\n            completed_tasks += 1\n    return completed_tasks",
        "rejected": "```python\nfrom collections import deque\n\ndef greedy_scheduling(tasks):\n    tasks = deque(sorted(tasks, key=lambda x: x[1]), maxlen=len(tasks))\n    completed = 0\n    current_deadline = 0\n\n    while tasks:\n        task, deadline = tasks.popleft()\n        if deadline >= current_deadline:\n            completed += 1\n            current_deadline = deadline\n\n    return completed\n\n# Example usage:\ntasks = [(2, 3), (1, 2), (6, 4), (8, 9), (4, 5)]\nresult = greedy_scheduling(tasks)\nprint(result)  # Output: 3\n```\n\n**Explanation:**\n\n* **Data Structures:**\n    * `deque` from collections module is used to maintain a doubly-ended queue for tasks, allowing for efficient insertion and deletion from both ends.\n* **Greedy Algorithm:**\n    * Tasks are sorted based on their deadlines in ascending order.\n    * The algorithm maintains two pointers: current deadline and number of completed tasks.\n    * It iterates through tasks and adds them to the completed count if their deadline is greater than or equal to the current deadline.\n    * The current deadline is updated to the deadline of the completed task.\n* **Output:**\n    * The function returns the maximum number of tasks that can be completed by the deadline.\n\n**Example Usage:**\n\n* The example input contains five tasks with deadlines.\n* The function returns `3`, which means that three tasks can be completed by their deadlines.\n* The results are printed to the console.\n\n**Time Complexity:** O(n log n), where n is the number of tasks.\n\n**Space Complexity:** O(n), due to the use of deque data structure.\n\n**Note:**\n\n* This algorithm assumes that tasks can only be completed by their deadlines.\n* It does not handle tasks with overlapping deadlines.\n* For more complex scheduling problems, other algorithms such as Dijkstra's algorithm or other heuristic approaches may be more suitable."
    },
    {
        "id": 36,
        "prompt": "Implement a Python function to solve the Activity Selection Problem using a greedy algorithm. The function should take a list of activities with their start and end times as input and return the maximum number of activities that can be performed by a single person.",
        "chosen": "def activity_selection_greedy(activities):\n    activities.sort(key=lambda x: x[1])\n    selected_activities = [activities[0]]\n    for activity in activities[1:]:\n        if activity[0] >= selected_activities[-1][1]:\n            selected_activities.append(activity)\n    return selected_activities",
        "rejected": "```python\ndef activity_selection(activities):\n    activities.sort(key=lambda x: x[1])  # Sort activities by end time\n\n    selected = []\n    current_end_time = float('-inf')\n\n    for start, end in activities:\n        if start >= current_end_time:\n            selected.append((start, end))\n            current_end_time = end\n\n    return len(selected)\n```\n\n**Explanation:**\n\n* **Sorting:** The activities are sorted in ascending order of their end times. This sorting criterion ensures that we can always select the next activity without violating the constraint that no activity can start before the previous one ends.\n\n\n* **Iterating over activities:** We iterate over each activity in the input list.\n\n\n* **Selection criterion:** If the current activity's start time is greater than or equal to the current end time of the previous activity, we select the current activity and update the current end time. This ensures that we are selecting non-overlapping activities.\n\n\n* **Output:** The function returns the maximum number of activities that can be selected without violating the constraints.\n\n**Example:**\n\n```python\nactivities = [\n    [2, 3],\n    [1, 2],\n    [3, 4],\n    [6, 8],\n]\n\nnum_activities = activity_selection(activities)\n\nprint(f\"Maximum number of activities: {num_activities}\")\n```\n\n**Output:**\n\n```\nMaximum number of activities: 3\n```\n\n**Time Complexity:** O(n log n), where n is the number of activities.\n\n**Space Complexity:** O(n), due to the creation of the sorted list and the selected activities list."
    },
    {
        "id": 37,
        "prompt": "Write a Python function to implement the Merge Sort algorithm.",
        "chosen": "def merge_sort(arr):\n    if len(arr) <= 1:\n        return arr\n    mid = len(arr) // 2\n    left_half = arr[:mid]\n    right_half = arr[mid:]\n    return merge(merge_sort(left_half), merge_sort(right_half))\n\ndef merge(left, right):\n    merged = []\n    left_index = 0\n    right_index = 0\n    while left_index < len(left) and right_index < len(right):\n        if left[left_index] <= right[right_index]:\n            merged.append(left[left_index])\n            left_index += 1\n        else:\n            merged.append(right[right_index])\n            right_index += 1\n    merged.extend(left[left_index:])\n    merged.extend(right[right_index:])\n    return merged",
        "rejected": "```python\ndef merge_sort(arr):\n    if len(arr) <= 1:\n        return arr\n\n    mid = len(arr) // 2\n    left_half = arr[:mid]\n    right_half = arr[mid:]\n\n    return merge(merge_sort(left_half), merge_sort(right_half))\n\ndef merge(left, right):\n    merged = []\n    left_index = 0\n    right_index = 0\n\n    while left_index < len(left) and right_index < len(right):\n        if left[left_index] <= right[right_index]:\n            merged.append(left[left_index])\n            left_index += 1\n        else:\n            merged.append(right[right_index])\n            right_index += 1\n\n    merged.extend(left[left_index:])\n    merged.extend(right[right_index:])\n\n    return merged\n```\n\n**Explanation:**\n\n* **merge_sort()** function:\n    * If the list has 0 or 1 element, it simply returns the list.\n    * Otherwise, it splits the list into two halves.\n    * It recursively calls itself on both halves to sort them.\n    * Finally, it merges the sorted halves using the **merge()** function.\n\n\n* **merge()** function:\n    * Creates an empty list to store the merged result.\n    * Indexes to track the current positions in the left and right lists.\n    * Iterates over both lists, comparing the elements.\n    * If the left list's element is less than or equal to the right list's element, it is appended to the merged list, and the left index is incremented.\n    * Otherwise, the right list's element is appended, and the right index is incremented.\n    * Once both lists are exhausted, the remaining elements from both lists are appended to the merged list.\n\n\n**Example Usage:**\n\n```python\n# Example list to sort\narr = [5, 2, 4, 6, 1, 3]\n\n# Sort the list using merge sort\nsorted_arr = merge_sort(arr)\n\n# Print the sorted list\nprint(sorted_arr)  # Output: [1, 2, 3, 4, 5, 6]\n```\n\n**Time Complexity:** O(n log n)\n**Space Complexity:** O(n)"
    },
    {
        "id": 38,
        "prompt": "How does the time complexity of the Merge Sort algorithm change when the input size is doubled?",
        "chosen": "The time complexity of the Merge Sort algorithm remains O(n log n) even when the input size is doubled, because the logarithmic term in the time complexity does not change with the size of the input.",
        "rejected": "**Time complexity of Merge Sort:**\n\n- Worst, Best, and Average case: O(n log n)\n\n**Impact of doubling the input size:**\n\nWhen the input size is doubled, the time complexity of Merge Sort increases by a factor of **log 2**.\n\n**Reason:**\n\n- Merge Sort consists of multiple recursive steps.\n- Each step involves dividing the input list into smaller and smaller sublists until each sublist contains only one element.\n- The depth of the recursion tree is O(log n), which means there are O(log n) levels of recursion.\n- Doubling the input size doubles the depth of the recursion tree, resulting in an additional factor of log 2 in the time complexity.\n\n**Therefore, the time complexity of Merge Sort changes from O(n log n) to O(2n log n) when the input size is doubled.**"
    },
    {
        "id": 39,
        "prompt": "Write a Python function to implement the Quick Sort algorithm.",
        "chosen": "def quick_sort(arr):\n    if len(arr) <= 1:\n        return arr\n    pivot = arr[len(arr) // 2]\n    left = [x for x in arr if x < pivot]\n    middle = [x for x in arr if x == pivot]\n    right = [x for x in arr if x > pivot]\n    return quick_sort(left) + middle + quick_sort(right)",
        "rejected": "```python\ndef quicksort(arr):\n    if len(arr) <= 1:\n        return arr\n\n    pivot = arr[0]\n    less = [x for x in arr[1:] if x < pivot]\n    greater = [x for x in arr[1:] if x >= pivot]\n\n    return quicksort(less) + [pivot] + quicksort(greater)\n```\n\n**Explanation:**\n\n* **Base Case:** If the list has 0 or 1 element, it is already sorted, so it is simply returned.\n\n\n* **Selection of Pivot:** The first element in the list is chosen as the pivot.\n\n\n* **Partitioning:** The remaining elements in the list are partitioned into two sets: those less than the pivot and those greater than or equal to the pivot.\n\n\n* **Sorting Recursively:** The sorting process is recursively repeated on the two partitions.\n\n\n**Example Usage:**\n\n```python\n# Example list to sort\narr = [5, 2, 4, 6, 1, 3]\n\n# Sort the list using quicksort\nsorted_arr = quicksort(arr)\n\n# Print the sorted list\nprint(sorted_arr)  # Output: [1, 2, 3, 4, 5, 6]\n```\n\n**Time Complexity:**\n\n* Best and Average Case: O(n log n)\n* Worst Case: O(n^2)\n\n**Space Complexity:**\n\n* O(log n)\n\n**Advantages of Quick Sort:**\n\n* Efficient for large datasets.\n* Stable sorting algorithm.\n* In-place sorting algorithm.\n\n**Disadvantages of Quick Sort:**\n\n* Worst-case time complexity can be O(n^2).\n* Not suitable for small datasets.\n* Can perform poorly when the pivot is not chosen carefully.\n\n**Note:**\n\n* The choice of pivot can affect the efficiency of the algorithm.\n* It is important to handle duplicates in the input list."
    },
    {
        "id": 40,
        "prompt": "What is the worst-case time complexity of the Quick Sort algorithm?",
        "chosen": "The worst-case time complexity of the Quick Sort algorithm is O(n^2), which occurs when the pivot is always the smallest or largest element in the array.",
        "rejected": "**Worst-case time complexity of Quick Sort:** **O(n^2)**\n\nQuick Sort is a divide-and-conquer sorting algorithm, but in the worst case, it can perform poorly, with a time complexity of O(n^2). This occurs when the chosen pivot is the smallest or largest element in the input list, leading to unbalanced partitioning and repeated selection of the same element as the pivot."
    },
    {
        "id": 41,
        "prompt": "Write a Python function to implement the Binary Search algorithm.",
        "chosen": "def binary_search(arr, target):\n    low, high = 0, len(arr) - 1\n    while low <= high:\n        mid = (low + high) // 2\n        if arr[mid] == target:\n            return mid\n        elif arr[mid] < target:\n            low = mid + 1\n        else:\n            high = mid - 1\n    return -1",
        "rejected": "```python\ndef binary_search(arr, target):\n    low = 0\n    high = len(arr) - 1\n\n    while low <= high:\n        mid = (low + high) // 2\n\n        if arr[mid] == target:\n            return mid\n        elif arr[mid] < target:\n            low = mid + 1\n        else:\n            high = mid - 1\n\n    return -1\n```\n\n**Explanation:**\n\n* **Initialization:**\n    * Set the search boundaries: `low` and `high` pointers.\n* **Loop:**\n    * While `low` is less than or equal to `high`:\n        * Calculate the midpoint `mid`.\n        * Check if the target element is equal to `arr[mid]`:\n            * If yes, return the index `mid`.\n        * If `arr[mid]` is less than the target, move the `low` pointer to the right.\n        * Otherwise, move the `high` pointer to the left.\n* **Termination:**\n    * When the `low` pointer becomes greater than the `high` pointer, the target element is not found, so return -1.\n* **Result:**\n    * The function returns the index of the target element if found, or -1 if it is not found.\n\n**Example Usage:**\n\n```python\n# Example list\narr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]\n\n# Target element to search for\ntarget = 5\n\n# Perform binary search\nresult = binary_search(arr, target)\n\n# Print the result\nprint(result)  # Output: 4\n```\n\n**Time Complexity:** O(log n), where n is the length of the list.\n\n**Space Complexity:** O(1)\n\n**Note:**\n\n* This function assumes that the input list `arr` is sorted in ascending order.\n* The `target` argument should be an integer representing the element to search for.\n* The function returns -1 if the target element is not found."
    },
    {
        "id": 42,
        "prompt": "What is the time complexity of the Binary Search algorithm?",
        "chosen": "The time complexity of the Binary Search algorithm is O(log n).",
        "rejected": "**Time complexity of the Binary Search algorithm:**\n\n**O(log n)**\n\n**Explanation:**\n\n* Binary Search is a search algorithm that efficiently locates an element in a sorted array.\n* The algorithm divides the search space in half at each iteration.\n* Since the array is sorted, the element can be found in the smaller half in O(log n) time.\n* Each comparison reduces the search space by half, resulting in a logarithmic number of comparisons to find the element.\n\n**The time complexity breakdown is as follows:**\n\n- **Input size:** n (size of the array)\n- **Number of comparisons:** O(log n)\n\n**Factors influencing the time complexity:**\n\n- The size of the array (n)\n- The number of elements in the array\n- The presence of duplicates\n\n**Example:**\n\n- Searching for an element in a sorted array of size 100 takes approximately 4 comparisons on average.\n- For an array of size 1000, it takes approximately 10 comparisons on average.\n\n**Advantages of Binary Search:**\n\n- Efficient search time complexity.\n- Suitable for large datasets.\n- Stable and reliable algorithm.\n\n**Disadvantages of Binary Search:**\n\n- Only works on sorted arrays.\n- Not suitable for searching unsorted datasets."
    },
    {
        "id": 43,
        "prompt": "Implement a recursive function to solve the N-Queens problem in Python.",
        "chosen": "def solve_n_queens(n):\n    def is_valid(board, row, col):\n        for i in range(row):\n            if board[i] == col or \\\n                board[i] - i == col - row or \\\n                board[i] + i == col + row:\n                return False\n        return True\n\n    def solve_n_queens_recursive(n, row, board):\n        if row == n:\n            result.append(board[:])\n            return\n        for col in range(n):\n            if is_valid(board, row, col):\n                board[row] = col\n                solve_n_queens_recursive(n, row + 1, board)\n\n    result = []\n    solve_n_queens_recursive(n, 0, [-1] * n)\n    return result",
        "rejected": "```python\ndef solveNQueens(n):\n    def isValid(board, row, col):\n        for i in range(row):\n            if board[i][col] == 1:\n                return False\n        for i in range(row, -1, -1):\n            if board[i][col - i] == 1 or board[i][col + i] == 1:\n                return False\n        return True\n\n    def backtrack(row, col):\n        if row == n:\n            return True\n        if col == n:\n            return backtrack(row + 1, 0)\n\n        for col in range(n):\n            if isValid(board, row, col):\n                board[row][col] = 1\n                if backtrack(row, col + 1):\n                    return True\n                board[row][col] = 0\n\n        return False\n\n    board = [[0 for _ in range(n)] for _ in range(n)]\n    return backtrack(0, 0)\n```\n\n**Explanation:**\n\n* **isValid()** function checks if a move is valid, ensuring that no other queen threatens the current position.\n* **backtrack()** function recursively explores the board, placing queens one at a time.\n* **row** and **col** represent the current row and column being considered.\n* **board** is a 2D list representing the chessboard, where 1s indicate the presence of a queen.\n\n**Example Usage:**\n\n```python\nn = 4\nresult = solveNQueens(n)\nprint(result)\n```\n\n**Output:**\n\n```\nTrue\n```\n\n**Note:**\n\n* The output indicates that a solution was found.\n* The `result` variable will be `True` if a solution is found, or `False` otherwise.\n* The solution is not explicitly returned, but it is stored in the `board` variable.\n\n**Time Complexity:**\n\n* O(4^n/sqrt(n))\n\n**Space Complexity:**\n\n* O(n)\n\n**Note:**\n\n* This implementation uses recursion, which can be inefficient for large values of n. For larger problems, iterative approaches are more suitable.\n* The code assumes that the board is a square grid."
    },
    {
        "id": 44,
        "prompt": "Write a Python function to solve the Sudoku puzzle using backtracking.",
        "chosen": "def is_valid(board, row, col, num):\n    for x in range(9):\n        if board[row][x] == num:\n            return False\n    for x in range(9):\n        if board[x][col] == num:\n            return False\n    start_row, start_col = row - row % 3, col - col % 3\n    for i in range(3):\n        for j in range(3):\n            if board[i + start_row][j + start_col] == num:\n                return False\n    return True\n\ndef solve_sudoku(board):\n    for i in range(9):\n        for j in range(9):\n            if board[i][j] == 0:\n                for num in range(1, 10):\n                    if is_valid(board, i, j, num):\n                        board[i][j] = num\n                        if solve_sudoku(board):\n                            return True\n                        board[i][j] = 0\n                return False\n        return True\n\n# Example usage:\nboard = [\n    [5, 3, 0, 0, 7, 0, 0, 0, 0],\n    [6, 0, 0, 1, 9, 5, 0, 0, 0],\n    [0, 9, 8, 0, 0, 0, 0, 6, 0],\n    [8, 0, 0, 0, 6, 0, 0, 0, 3],\n    [4, 0, 0, 8, 0, 3, 0, 0, 1],\n    [7, 0, 0, 0, 2, 0, 0, 0, 6],\n    [0, 6, 0, 0, 0, 0, 2, 8, 0],\n    [0, 0, 0, 4, 1, 9, 0, 0, 5],\n    [0, 0, 0, 0, 8, 0, 0, 7, 9]\n]\nsolve_sudoku(board)\nprint(board)",
        "rejected": "```python\ndef solve_sudoku(puzzle):\n    def is_valid(row, col, num):\n        for i in range(9):\n            if puzzle[row][i] != 0 and i != col:\n                if puzzle[row][i] == num:\n                    return False\n        for i in range(9):\n            if puzzle[i][col] != 0 and i != row:\n                if puzzle[i][col] == num:\n                    return False\n        return True\n\n    def backtrack(row, col):\n        if row == 9 and col == 9:\n            return True\n        if col == 9:\n            return backtrack(row + 1, 0)\n        if puzzle[row][col] != 0:\n            return backtrack(row, col + 1)\n        for num in range(1, 10):\n            if is_valid(row, col, num):\n                puzzle[row][col] = num\n                if backtrack(row, col + 1):\n                    return True\n                puzzle[row][col] = 0\n        return False\n\n    return backtrack(0, 0)\n\n```\n\n**Usage:**\n\n```python\n# Example Sudoku puzzle\npuzzle = [\n    [0, 5, 0, 0, 2, 0, 0, 0, 0],\n    [6, 0, 0, 1, 9, 5, 0, 0, 0],\n    [0, 9, 8, 0, 0, 0, 0, 6, 0],\n    [8, 0, 0, 0, 6, 0, 0, 0, 3],\n    [4, 0, 0, 8, 0, 3, 0, 0, 1],\n    [7, 0, 0, 0, 0, 0, 2, 8, 0],\n    [0, 6, 0, 0, 0, 0, 0, 7, 9],\n    [0, 0, 0, 4, 1, 9, 0, 0, 5],\n    [0, 0, 0, 0, 8, 0, 0, 0, 6],\n]\n\n# Solve the Sudoku puzzle\nsolution = solve_sudoku(puzzle)\n\n# Print the solution\nprint(solution)\n```\n\n**Output:**\n\n```\n[[5, 5, 4, 7, 2, 6, 9, 1, 8],\n [6, 1, 3, 1, 9, 5, 8, 4, 2],\n [2, 9, 8, 3, 7, 4, 5, 6, 3],\n [8, 3, 7, 9, 6, 1, 4, 2, 3],\n [4, 2, 9, 8, 5, 3, 7, 3, 1],\n [7, 4, 1, 5, 3, 8, 2, 8, 6],\n [1, 6, 5, 2, 4, 7, 3, 7, 9],\n [9, 8, 6, 4, 1, 9, 6, 5, 5],\n [3, 7, 2, 6, 8, 5, 1, 9, 6]]\n```\n\n**Note:**\n\n* This function uses backtracking algorithm to solve the Sudoku puzzle.\n* The `is_valid()` function checks if a number is valid for a given row and column.\n* The `backtrack()` function recursively tries different numbers for each cell until it finds a solution.\n* The `solve_sudoku()` function calls the `backtrack()` function to solve the puzzle."
    },
    {
        "id": 45,
        "prompt": "Explain the concept of backtracking in the context of algorithms and provide a Python implementation of the Rat in a Maze problem using backtracking.",
        "chosen": "Backtracking is an algorithmic-technique for solving problems recursively by trying to build a solution incrementally, one piece at a time, removing those solutions that fail to satisfy the constraints of the problem at any point of time (by time, here, is referred to the time elapsed till reaching any level of the search tree). \n\nPython implementation of the Rat in a Maze problem using backtracking:\ndef is_valid_move(maze, x, y):\n    return 0 <= x < len(maze) and 0 <= y < len(maze[0]) and maze[x][y] == 1\n\ndef solve_maze(maze, x, y, path):\n    if x == len(maze) - 1 and y == len(maze[0]) - 1:\n        return True\n    if not is_valid_move(maze, x, y):\n        return False\n    maze[x][y] = 0\n    path.append((x, y))\n    for dx, dy in [(-1, 0), (1, 0), (0, -1), (0, 1)]:\n        if solve_maze(maze, x + dx, y + dy, path):\n            return True\n    path.pop()\n    return False\n\n# Example usage:\nmaze = [\n    [1, 0, 0, 0, 1],\n    [1, 1, 1, 1, 1],\n    [0, 1, 0, 0, 0],\n    [1, 1, 0, 1, 1],\n    [1, 1, 1, 1, 1]\n]\npath = []\nif solve_maze(maze, 0, 0, path):\n    print(path)\nelse:\n    print(\"No path found\")",
        "rejected": "## Backtracking\n\n**Concept:**\n\nBacktracking is a search algorithm technique where a problem is solved by systematically exploring potential solutions and reversing or retracting steps when necessary. It involves:\n\n1. Exploring a solution space by making choices.\n2. Checking if the current solution is valid.\n3. If valid, proceed. If not, backtrack and try a different choice.\n4. Continue exploring until a solution is found or all possibilities have been exhausted.\n\n\n**Rat in a Maze Problem:**\n\nThe Rat in a Maze problem involves finding a path from the starting point to the exit in a maze without obstacles.\n\n**Python Implementation:**\n\n```python\ndef rat_maze(maze, start, end):\n    if start == end:\n        return True\n\n    for neighbor in get_neighbors(start):\n        if is_valid(maze, neighbor):\n            if rat_maze(maze, neighbor, end):\n                return True\n\n    return False\n\n# Other functions:\n- get_neighbors: Calculates neighboring cells\n- is_valid: Checks if a cell is valid (not an obstacle)\n```\n\n**Explanation:**\n\n1. The function takes three arguments: the maze, the start and end points.\n2. It starts by checking if the start point is the end point. If yes, it returns True.\n3. It iterates over neighboring cells of the current position.\n4. If the neighbor is valid (not an obstacle), it recursively calls itself with the neighbor as the new start point.\n5. If the recursion returns True, it means a path was found.\n6. If the recursion returns False, it means no path was found from that neighbor.\n7. If no valid neighbors are found, the function returns False.\n\n**Key Aspects:**\n\n- The solution involves exploring multiple paths until a valid path is found.\n- The algorithm backtracks when it encounters an obstacle or reaches a dead end.\n- The recursion depth is limited by the size of the maze.\n\n**Time Complexity:**\n\n- Worst case: O(2^n), where n is the size of the maze.\n- Best case: O(n), when the solution is found quickly.\n\n**Space Complexity:**\n\n- O(n), due to the recursive call stack."
    },
    {
        "id": 46,
        "prompt": "Implement a Python function to solve the subset sum problem using backtracking.",
        "chosen": "def is_subset_sum(set, n, sum):\n    if sum == 0:\n        return True\n    if n == 0 and sum != 0:\n        return False\n    if set[n-1] > sum:\n        return is_subset_sum(set, n-1, sum)\n    return is_subset_sum(set, n-1, sum) or is_subset_sum(set, n-1, sum-set[n-1])\n\n# Example usage:\nset = [3, 9, 8, 4, 5, 7]\nsum = 15\nn = len(set)\nif is_subset_sum(set, n, sum):\n    print(\"Subset sum exists\")\nelse:\n    print(\"Subset sum does not exist\")",
        "rejected": "```python\ndef subset_sum(nums, target):\n    def backtrack(index, current, remaining):\n        if remaining == 0:\n            return True\n        if remaining < 0:\n            return False\n        if index == len(nums):\n            return False\n\n        for i in range(index, len(nums)):\n            current += nums[i]\n            if backtrack(i + 1, current, remaining - nums[i]):\n                return True\n            current -= nums[i]\n\n    return backtrack(0, 0, target)\n```\n\n**How it works:**\n\n* **Backtracking algorithm:**\n    - The function starts by selecting an index `index` from the input list `nums`.\n    - `current` keeps track of the sum of the current subset.\n    - `remaining` is the remaining target sum.\n* **Base cases:**\n    - If `remaining` becomes 0, we have found a subset that sums up to the target, and the function returns `True`.\n    - If `remaining` becomes negative, we know that we cannot create a subset that sums up to the target, and the function returns `False`.\n* **Recursive step:**\n    - For each index `i` from `index` to the end of the list, we:\n        - Add `nums[i]` to `current`.\n        - Call the `backtrack` function recursively with the updated `current` and `remaining - nums[i]`.\n        - If a subset is found, the function returns `True`.\n        - If no subset is found, we subtract `nums[i]` from `current`.\n\n**Example:**\n\n```python\nnums = [2, 3, 6, 7, 8, 10]\ntarget = 7\n\nresult = subset_sum(nums, target)\n\nprint(result)  # Output: True\n```\n\n**Time complexity:** O(2^n), where n is the length of the input list `nums`.\n\n**Space complexity:** O(1)\n\n**Note:** This is a recursive solution, which can be inefficient for large input sizes. For larger problems, it is recommended to use iterative approaches."
    },
    {
        "id": 47,
        "prompt": "Write a Python function to solve the N-Queens problem using backtracking.",
        "chosen": "def solve_n_queens(n):\n    def is_valid(board, row, col):\n        for i in range(row):\n            if board[i] == col or \\\n                board[i] - i == col - row or \\\n                board[i] + i == col + row:\n                return False\n        return True\n\n    def solve_n_queens_recursive(n, row, board):\n        if row == n:\n            result.append(board[:])\n            return\n        for col in range(n):\n            if is_valid(board, row, col):\n                board[row] = col\n                solve_n_queens_recursive(n, row + 1, board)\n\n    result = []\n    solve_n_queens_recursive(n, 0, [-1] * n)\n    return result\n\n# Example usage:\nn = 4\nresult = solve_n_queens(n)\nfor solution in result:\n    print(solution)",
        "rejected": "```python\ndef solveNQueens(n):\n    def isValid(board, row, col):\n        for i in range(row):\n            if board[i][col] == 1:\n                return False\n        for i in range(col):\n            if board[row][i] == 1:\n                return False\n        return True\n\n    def backtrack(row, col, board):\n        if row == n:\n            return True\n        if col == n:\n            return backtrack(row+1, 0, board)\n\n        for col in range(n):\n            if isValid(board, row, col):\n                board[row][col] = 1\n                if backtrack(row, col+1, board):\n                    return True\n                board[row][col] = 0\n\n        return False\n\n    board = [[0 for _ in range(n)] for _ in range(n)]\n    return backtrack(0, 0, board)\n```\n\n**Usage:**\n\n```python\nn = 4\nresult = solveNQueens(n)\n\nif result:\n    for row in result:\n        print(row)\nelse:\n    print(\"No solution found\")\n```\n\n**Output:**\n\n```\n[1, 0, 0, 0]\n[0, 1, 0, 0]\n[0, 0, 1, 0]\n[0, 0, 0, 1]\n```\n\n**Explanation:**\n\n* The `solveNQueens` function takes the size of the board (n) as input.\n* The `isValid` function checks if a move is valid, ensuring that no other queen threatens the current position.\n* The `backtrack` function recursively explores the board, placing queens one at a time and checking if a solution is found.\n* If a solution is found, it is returned. Otherwise, the function returns `False`.\n\n**Note:**\n\n* The solution returned by the function is a list of lists, where each inner list represents a row of the board.\n* The value `1` in the board represents a queen, while `0` represents an empty cell."
    },
    {
        "id": 48,
        "prompt": "Explain the concept of backtracking in the context of algorithms and provide a Python implementation of the Knight's Tour problem using backtracking.",
        "chosen": "Backtracking is an algorithmic-technique for solving problems recursively by trying to build a solution incrementally, one piece at a time, removing those solutions that fail to satisfy the constraints of the problem at any point of time (by time, here, is referred to the time elapsed till reaching any level of the search tree). \n\nPython implementation of the Knight's Tour problem using backtracking:\ndef is_valid_move(board, x, y, moves, move_num):\n    if 0 <= x < len(board) and 0 <= y < len(board[0]) and board[x][y] == -1:\n        return True\n    return False\n\ndef solve_knights_tour(board, x, y, moves, move_num):\n    if move_num == len(board) * len(board[0]):\n        return True\n    for dx, dy in moves:\n        nx, ny = x + dx, y + dy\n        if is_valid_move(board, nx, ny, moves, move_num):\n            board[nx][ny] = move_num\n            if solve_knights_tour(board, nx, ny, moves, move_num + 1):\n                return True\n            board[nx][ny] = -1\n    return False\n\n# Example usage:\nboard = [[-1]*8 for _ in range(8)]\nmoves = [(2, 1), (1, 2), (-1, 2), (-2, 1), (-2, -1), (-1, -2), (1, -2), (2, -1)]\nboard[0][0] = 0\nif solve_knights_tour(board, 0, 0, moves, 1):\n    for row in board:\n        print(row)\nelse:\n    print(\"No solution exists\")",
        "rejected": "## Backtracking\n\nBacktracking is a problem-solving paradigm that involves exploring potential solutions by repeatedly expanding and retracting steps until a solution is found. It involves:\n\n**1. Exploring:**\n- Starting with a partial solution (initial state).\n- Expanding the solution by making valid moves.\n\n\n**2. Checking:**\n- Checking if the solution is complete (goal state).\n- If yes, the solution is accepted.\n- If no, backtrack by retracting the last move.\n\n\n**3. Expanding:**\n- If the solution is not complete, explore other possible moves from the current state.\n\n\n**4. Repeating:**\n- Steps 2 and 3 are repeated until either a solution is found or all possible moves have been explored.\n\n\n**Applications of Backtracking:**\n\n- Maze solving\n- Crossword puzzles\n- Scheduling problems\n- Path planning\n- Knight's Tour\n\n\n## Knight's Tour Problem\n\nThe Knight's Tour problem involves finding a path for a knight to visit every square of a chessboard once and only once.\n\n\n**Python Implementation:**\n\n```python\ndef is_valid(board, row, col):\n    return row >= 0 and row < len(board) and col >= 0 and col < len(board[0]) and board[row][col] == 0\n\ndef knight_tour(board, row, col, path):\n    if len(path) == len(board) * len(board[0]):\n        return True\n\n    for row_move, col_move in [(1, 2), (2, 1), (2, -1), (1, -2), (-1, -2), (-2, -1), (-2, 1), (-1, 2)]:\n        new_row, new_col = row + row_move, col + col_move\n        if is_valid(board, new_row, new_col):\n            board[new_row][new_col] = len(path) + 1\n            if knight_tour(board, new_row, new_col, path + [new_row, new_col]):\n                return True\n    board[row][col] = 0\n    return False\n\ndef solve_knight_tour(board):\n    if knight_tour(board, 0, 0, []):\n        return True\n    else:\n        return False\n```\n\n**How it works:**\n\n- The `knight_tour` function takes the current row and column, along with a list of visited squares.\n- It checks if the current position has already been visited.\n- If not, it iterates over possible moves of the knight and recursively calls itself with the updated position and visited list.\n- If a solution is found, it returns `True`.\n- If no solution is found, it returns `False`.\n\n\n**Note:**\n\n- The provided code is a recursive implementation.\n- Recursive algorithms can be memory inefficient for large problems. In such cases, iterative approaches are more suitable."
    },
    {
        "id": 49,
        "prompt": "Implement a Python function to solve the Hamiltonian Cycle problem using backtracking.",
        "chosen": "def is_valid_move(graph, path, v):\n    if v not in path:\n        return True\n    return False\n\ndef solve_hamiltonian_cycle(graph, path, v):\n    if len(path) == len(graph):\n        if graph[path[-1]][path[0]] == 1:\n            return True\n        return False\n    for i in range(len(graph)):\n        if is_valid_move(graph, path, i):\n            path.append(i)\n            if solve_hamiltonian_cycle(graph, path, i):\n                return True\n            path.pop()\n    return False\n\n# Example usage:\ngraph = [[0, 1, 1, 1, 0, 0, 0, 0, 0, 0],\n         [1, 0, 1, 0, 0, 0, 0, 0, 0, 0],\n         [1, 1, 0, 1, 0, 0, 0, 0, 0, 0],\n         [1, 0, 1, 0, 1, 0, 0, 0, 0, 0],\n         [0, 0, 0, 1, 0, 1, 0, 0, 0, 0],\n         [0, 0, 0, 0, 1, 0, 1, 0, 0, 0],\n         [0, 0, 0, 0, 0, 1, 0, 1, 0, 0],\n         [0, 0, 0, 0, 0, 0, 1, 0, 1, 0],\n         [0, 0, 0, 0, 0, 0, 0, 1, 0, 1],\n         [0, 0, 0, 0, 0, 0, 0, 0, 1, 0],\n         [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]\npath = [0]\nif solve_hamiltonian_cycle(graph, path, 0):\n    print(path)\nelse:\n    print(\"No Hamiltonian cycle exists\")",
        "rejected": "```python\nfrom collections import defaultdict\n\ndef hamiltonian_cycle(graph):\n    \"\"\"\n    Solves the Hamiltonian Cycle problem using backtracking.\n\n    Args:\n        graph: A dictionary representing the graph, where keys are nodes and values are lists of neighboring nodes.\n\n    Returns:\n        A list of nodes representing a Hamiltonian cycle, or None if no such cycle exists.\n    \"\"\"\n\n    def backtrack(visited, current, previous):\n        if len(visited) == n:\n            if current == start and previous != start:\n                return visited\n\n        for neighbor in graph[current] - {previous}:\n            if neighbor not in visited:\n                visited = visited + [neighbor]\n                result = backtrack(visited, neighbor, current)\n                if result is not None:\n                    return result\n\n        return None\n\n    n = len(graph)\n    start = 0\n    visited = [False] * n\n    result = backtrack(visited, start, None)\n    return result\n```\n\n**Explanation:**\n\n* **Backtracking Algorithm:**\n    * The function starts by initializing a list `visited` to track the current traversal state.\n    * It iteratively visits neighboring nodes of the current node until it completes a cycle.\n    * If the end node is the same as the start node and the previous node is not the start node, a Hamiltonian cycle has been found.\n    * If no cycle is found, the function returns `None`.\n\n\n* **Graph Representation:**\n    * The graph is represented as a dictionary, where keys are nodes and values are lists of neighboring nodes.\n\n\n* **Complexity:**\n    * Time Complexity: O(n * n), where n is the number of nodes.\n    * Space Complexity: O(n), due to the use of `visited` list.\n\n\n**Example:**\n\n```python\n# Example graph\ngraph = {\n    'A': ['B', 'C'],\n    'B': ['C', 'D'],\n    'C': ['D', 'A'],\n    'D': ['B'],\n}\n\n# Find a Hamiltonian cycle\ncycle = hamiltonian_cycle(graph)\n\n# Print result\nprint(cycle)  # Output: ['A', 'B', 'C', 'D']\n```\n\n**Note:**\n\n* This function assumes that the graph is undirected.\n* It can handle cycles in the graph.\n* The start node is chosen as node `0`, but this can be adjusted."
    },
    {
        "id": 50,
        "prompt": "Write a Python function to implement a max heap and perform insertion and deletion operations.",
        "chosen": "```\ndef max_heapify(arr, n, i):\n    largest = i\n    left = 2 * i + 1\n    right = 2 * i + 2\n    if left < n and arr[i] < arr[left]:\n        largest = left\n    if right < n and arr[largest] < arr[right]:\n        largest = right\n    if largest != i:\n        arr[i], arr[largest] = arr[largest], arr[i]\n        max_heapify(arr, n, largest)\ndef insert_max_heap(arr, key):\n    n = len(arr)\n    arr.append(key)\n    i = n\n    while i != 0 and arr[(i-1)//2] < key:\n        arr[i] = arr[(i-1)//2]\n        i = (i-1)//2\n    arr[i] = key\ndef delete_max_heap(arr):\n    n = len(arr)\n    if n == 0:\n        return\n    temp = arr[0]\n    arr[0] = arr[n-1]\n    arr.pop()\n    max_heapify(arr, n-1, 0)\n```",
        "rejected": "```python\nimport random\n\nclass MaxHeap:\n    def __init__(self):\n        self.heap = []\n\n    def size(self):\n        return len(self.heap)\n\n    def is_empty(self):\n        return self.size() == 0\n\n    def parent(self, i):\n        return (i - 1) // 2\n\n    def left(self, i):\n        return 2 * i + 1\n\n    def right(self, i):\n        return 2 * i + 2\n\n    def max_heapify(self, i):\n        while i > 0:\n            parent_index = self.parent(i)\n            if self.heap[i] < self.heap[parent_index]:\n                self.heap[i], self.heap[parent_index] = self.heap[parent_index], self.heap[i]\n                i = parent_index\n\n    def insert(self, value):\n        self.heap.append(value)\n        i = self.size() - 1\n        while i > 0:\n            parent_index = self.parent(i)\n            if self.heap[i] > self.heap[parent_index]:\n                self.heap[i], self.heap[parent_index] = self.heap[parent_index], self.heap[i]\n                i = parent_index\n\n    def extract(self):\n        if self.is_empty():\n            return None\n        max_value = self.heap[0]\n        self.heap[0] = self.heap[-1]\n        del self.heap[-1]\n        i = 0\n        while i < self.size():\n            l = self.left(i)\n            r = self.right(i)\n            if l < self.size() and self.heap[l] > self.heap[i]:\n                i = l\n            elif r < self.size() and self.heap[r] > self.heap[i]:\n                i = r\n            else:\n                break\n            if self.heap[i] < self.heap[parent]:\n                self.heap[i], self.heap[parent] = self.heap[parent], self.heap[i]\n                i = parent\n        return max_value\n\n\n# Example usage:\nheap = MaxHeap()\nheap.insert(20)\nheap.insert(10)\nheap.insert(30)\nheap.insert(5)\n\nwhile not heap.is_empty():\n    max_value = heap.extract()\n    print(f\"Extracted maximum: {max_value}\")\n```\n\n**Time Complexity:**\n\n* Insert: O(log n)\n* Delete: O(log n)\n\n**Space Complexity:** O(n), where n is the size of the heap.\n\n**Note:**\n\n* The `extract()` operation removes the maximum value from the heap.\n* The heap is implemented using a list.\n* The `max_heapify()` method restores the heap property by swapping values up the tree.\n* The `insert()` and `extract()` operations take O(log n) time complexity, where n is the size of the heap."
    },
    {
        "id": 51,
        "prompt": "How do you implement a Trie data structure in Python?",
        "chosen": "```\nclass TrieNode:\n    def __init__(self):\n        self.children = {}\n        self.is_end_of_word = False\n\nclass Trie:\n    def __init__(self):\n        self.root = TrieNode()\n    def insert(self, word):\n        node = self.root\n        for char in word:\n            if char not in node.children:\n                node.children[char] = TrieNode()\n            node = node.children[char]\n        node.is_end_of_word = True\n    def search(self, word):\n        node = self.root\n        for char in word:\n            if char not in node.children:\n                return False\n            node = node.children[char]\n        return node.is_end_of_word\n    def starts_with(self, prefix):\n        node = self.root\n        for char in prefix:\n            if char not in node.children:\n                return False\n            node = node.children[char]\n        return True\n```",
        "rejected": "```python\nclass Node:\n    def __init__(self, character, is_terminal):\n        self.character = character\n        self.is_terminal = is_terminal\n        self.children = {}\n\n\nclass Trie:\n    def __init__(self):\n        self.root = Node('\\x00', False)\n\n    def insert(self, word):\n        current = self.root\n        for letter in word:\n            if letter not in current.children:\n                current.children[letter] = Node(letter, False)\n            current = current.children[letter]\n        current.is_terminal = True\n\n    def search(self, word):\n        current = self.root\n        for letter in word:\n            if letter not in current.children:\n                return False\n            current = current.children[letter]\n        return current.is_terminal\n```\n\n**Explanation:**\n\n* **Node** class: Represents a node in the Trie data structure, containing a character and a flag indicating if it is a terminal node.\n* **Trie** class: Implements a Trie data structure with a root node.\n* **Insert** method: Inserts a word into the Trie.\n* **Search** method: Searches for a word in the Trie and returns True or False.\n\n**Example Usage:**\n\n```python\ntrie = Trie()\ntrie.insert('apple')\ntrie.insert('banana')\ntrie.insert('cherry')\n\nis_present = trie.search('apple')  # True\n```\n\n**Time Complexity:**\n\n* Insert: O(m), where m is the length of the word.\n* Search: O(m), where m is the length of the word.\n\n**Space Complexity:** O(m), due to the storage of the Trie itself.\n\n**Key Features:**\n\n* Case-insensitive search.\n* Efficient insertion and search operations.\n* Compact memory usage for common prefixes.\n\n**Applications:**\n\n* Word completion suggestions.\n* Autocompletion in text editors.\n* Search engines.\n* Spell checkers.\n\n**Advantages of Trie Data Structure:**\n\n* Fast and efficient search performance.\n* Compact memory usage for common prefixes.\n* Suitable for large vocabularies.\n\n**Disadvantages of Trie Data Structure:**\n\n* Not suitable for storing very large vocabularies due to memory limitations.\n* Can become unbalanced for certain insertion patterns."
    },
    {
        "id": 52,
        "prompt": "Write a Python function to find all words in a Trie that start with a given prefix.",
        "chosen": "```\ndef find_words_with_prefix(trie, prefix):\n    node = trie.root\n    for char in prefix:\n        if char not in node.children:\n            return []\n        node = node.children[char]\n    return find_all_words(node, prefix)\ndef find_all_words(node, prefix):\n    words = []\n    if node.is_end_of_word:\n        words.append(prefix)\n    for char, child in node.children.items():\n        words.extend(find_all_words(child, prefix + char))\n    return words\n```",
        "rejected": "```python\nfrom typing import List\n\nfrom collections import defaultdict\n\ndef find_words_with_prefix(trie: 'Trie', prefix: str) -> List[str]:\n    \"\"\"\n    Finds all words in a Trie that start with a given prefix.\n\n    Args:\n        trie: A Trie object.\n        prefix: The prefix to search for.\n\n    Returns:\n        A list of words in the Trie that start with the given prefix.\n    \"\"\"\n\n    results = []\n    current = trie.root\n\n    for char in prefix:\n        if char not in current:\n            break\n        current = current[char]\n\n    if current is not None:\n        for word in current.values():\n            results.append(word)\n\n    return results\n```\n\n**Explanation:**\n\n* The function takes two arguments: a Trie object and a prefix.\n* It starts by initializing an empty list to store the results.\n* It iterates over the characters in the prefix.\n* If the current node in the Trie does not have the current character, it breaks out of the loop.\n* Otherwise, it moves to the child node for the current character.\n* If the end of the prefix is reached, it iterates over the values (words) of the current node and adds them to the results list.\n* Finally, the function returns the list of results.\n\n**Example:**\n\n```python\n# Example Trie object\ntrie = Trie()\ntrie.insert('ab')\ntrie.insert('abc')\ntrie.insert('abca')\ntrie.insert('abcd')\n\n# Find words starting with 'ab'\nresults = find_words_with_prefix(trie, 'ab')\nprint(results)  # Output: ['ab', 'abc', 'abca']\n```\n\n**Time Complexity:** O(len(prefix)), where len(prefix) is the length of the prefix.\n\n**Space Complexity:** O(len(trie)), where len(trie) is the number of nodes in the Trie."
    },
    {
        "id": 53,
        "prompt": "How do you implement a priority queue using a heap in Python?",
        "chosen": "```\nimport heapq\n\nclass PriorityQueue:\n    def __init__(self):\n        self._queue = []\n        self._index = 0\n    def is_empty(self):\n        return not self._queue\n    def push(self, item, priority):\n        heapq.heappush(self._queue, (-priority, self._index, item))\n        self._index += 1\n    def pop(self):\n        return heapq.heappop(self._queue)[-1]\n```",
        "rejected": "**Step 1: Choose a Data Structure**\n\n* A heap is a suitable data structure for implementing a priority queue.\n* Python provides the `heap` module with built-in functions for heap operations.\n\n\n**Step 2: Defining the Priority Queue**\n\n```python\nimport heap\n\ndef priority_queue():\n    return []\n```\n\n**Step 3: Inserting an Element**\n\n```python\ndef enqueue(queue, item, priority):\n    heap.heappush(queue, (priority, item))\n```\n\n**Step 4: Removing the Highest Priority Element**\n\n```python\ndef dequeue(queue):\n    return heap.heappop(queue)[1]\n```\n\n**Step 5: Checking if the Queue is Empty**\n\n```python\ndef is_empty(queue):\n    return not queue\n```\n\n**Example:**\n\n```python\n# Create a priority queue\nqueue = priority_queue()\n\n# Insert elements with priorities\nenqueue(queue, 'A', 1)\nenqueue(queue, 'B', 2)\nenqueue(queue, 'C', 3)\n\n# Dequeue the elements in order of priority\nwhile not is_empty(queue):\n    item = dequeue(queue)\n    print(item)  # Output: 'A', 'B', 'C'\n```\n\n**Time Complexity:**\n\n* Insertion: O(log n)\n* Dequeuing: O(log n)\n* Checking if empty: O(1)\n\n**Space Complexity:** O(n), where n is the size of the heap.\n\n**Advantages of Using a Heap for a Priority Queue:**\n\n* Efficient insertion and deletion of the highest priority element.\n* Stable and well-tested data structure.\n* Provides a complete binary tree structure.\n\n\n**Disadvantages of Using a Heap for a Priority Queue:**\n\n* Not suitable for storing large objects, as the heap can become memory inefficient.\n* Other data structures may be more efficient for certain operations, such as sorting or searching.\n\n**Note:**\n\n* The `heapify()` function can be used to convert any list into a heap.\n* The `heappop()` function removes and returns the highest priority element from the heap.\n* The `heapreplace()` function can be used to update the priority of an existing element in the heap."
    },
    {
        "id": 54,
        "prompt": "Write a Python function to find the kth largest element in an array using a heap.",
        "chosen": "```\ndef find_kth_largest(nums, k):\n    if k < 1 or k > len(nums):\n        return None\n    heap = []\n    for num in nums:\n        if len(heap) < k:\n            heapq.heappush(heap, num)\n        else:\n            heapq.heappushpop(heap, num)\n    return heap[0]\n```",
        "rejected": "```python\nimport heapq\n\ndef kth_largest(arr, k):\n    if len(arr) < k:\n        raise ValueError(\"k must be less than the length of the array\")\n\n    # Build a max heap to store the k largest elements\n    max_heap = [-x for x in arr]\n    heapify(max_heap)\n\n    # Extract the k largest elements one at a time\n    for i in range(k):\n        x = heapq.heappop(max_heap)\n\n    # Return the kth largest element\n    return -x\n```\n\n**Usage:**\n\n```python\n# Example array\narr = [5, 3, 8, 2, 10, 9, 4, 7, 6]\n\n# Find the 3rd largest element\nk = 3\nlargest = kth_largest(arr, k)\n\n# Print the result\nprint(largest)  # Output: 7\n```\n\n**Explanation:**\n\n* The function takes two arguments: `arr` (a list of numbers) and `k` (the index of the kth largest element to find).\n* It first checks if `k` is less than the length of the array. If not, it raises an error.\n* It then builds a max heap using the `heapify()` function from the `heapq` module. The heap will store the k largest elements in the array.\n* It iterates through the `k` elements and extracts the largest element from the heap using `heappop()`.\n* Finally, it returns the kth largest element, which is the top element of the heap.\n\n**Time Complexity:** O(k log k), where k is the value of `k`.\n\n**Space Complexity:** O(k), due to the creation of the heap.\n\n**Note:**\n\n* The `heapify()` function builds a heap from a list.\n* The `heappop()` function removes the top element (the largest element) from a heap.\n* The negative sign in the heap is used to maintain the max heap property. The actual values in the heap are negative of the actual values in the array."
    },
    {
        "id": 55,
        "prompt": "Implement a Trie-based autocomplete system in Python.",
        "chosen": "```\nimport collections\n\nclass TrieNode:\n    def __init__(self):\n        self.children = collections.defaultdict(TrieNode)\n        self.is_end_of_word = False\n\nclass AutocompleteSystem:\n    def __init__(self, sentences, times):\n        self.root = TrieNode()\n        for sentence, time in zip(sentences, times):\n            self._insert(sentence, time)\n    def _insert(self, sentence, time):\n        node = self.root\n        for char in sentence:\n            node = node.children[char]\n        node.is_end_of_word = True\n        node.time = time\n    def input(self, c):\n        if c != '#':\n            self.buffer += c\n            node = self.root\n            for char in self.buffer:\n                if char not in node.children:\n                    return []\n                node = node.children[char]\n            return self._get_sentences(node)\n        else:\n            return []\n    def _get_sentences(self, node):\n        sentences = []\n        self._get_sentences_helper(node, self.buffer, sentences)\n        return sentences\n    def _get_sentences_helper(self, node, sentence, sentences):\n        if node.is_end_of_word:\n            sentences.append((sentence, node.time))\n        for char, child in node.children.items():\n            self._get_sentences_helper(child, sentence + char, sentences)\n```",
        "rejected": "```python\nimport collections\n\nclass Node:\n    def __init__(self, char, is_end=False):\n        self.char = char\n        self.is_end = is_end\n        self.children = collections.defaultdict(Node)\n\n\nclass Trie:\n    def __init__(self):\n        self.root = Node('\\x00')\n\n    def insert(self, word):\n        current = self.root\n        for char in word:\n            current = current.children[char]\n            current.children[char] = Node(char)\n        current.children[char].is_end = True\n\n    def search(self, prefix):\n        current = self.root\n        for char in prefix:\n            if char not in current.children:\n                return []\n            current = current.children[char]\n        if current.is_end:\n            return [word for word in current.children]\n        else:\n            return []\n\n\n# Example Usage:\ntrie = Trie()\ntrie.insert('apple')\ntrie.insert('banana')\ntrie.insert('cherry')\n\nresults = trie.search('app')\nprint(results)  # Output: ['apple']\n```\n\n**Explanation:**\n\n* **Node** class: Represents a node in the Trie data structure, which contains a character and a boolean indicating if it marks the end of a word.\n\n\n* **Trie** class: Implements a Trie data structure with a root node and a dictionary of children nodes.\n\n\n* **insert()** method: Inserts a word into the Trie.\n\n\n* **search()** method: Searches the Trie for a prefix and returns a list of words that match the prefix.\n\n\n* **Example Usage:** Creates a Trie object and inserts three words: \"apple\", \"banana\", and \"cherry\". Then, it searches for the prefix \"app\" and prints the results.\n\n**Time Complexity:**\n\n* Insert: O(m), where m is the length of the word\n* Search: O(m), where m is the length of the prefix\n\n**Space Complexity:** O(m), due to the storage of the Trie itself.\n\n**Features:**\n\n* Case-insensitive search\n* Efficient prefix search\n* Compact data structure\n\n**Applications:**\n\n* Autocompletion in search engines\n* Code completion in IDEs\n* Spell checkers\n* Predictive text suggestions"
    }
]