LC-1561 C++ JS TS accepted

LC-1561/LC-1561.cpp

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+#include <vector>
+#include <algorithm>
+
+using namespace std;
+
+class Solution {
+public:
+    /**
+     * Concepts: Sorting
+     */
+    int maxCoins(vector<int>& piles) {
+        int pilesLen = piles.size();
+        if (pilesLen < 3) return 0;
+
+        sort(piles.begin(), piles.end(), compare);
+
+        int result = 0;
+        for (int i = 1; i < 2 * pilesLen / 3; i += 2) {
+            result += piles[i];
+        }
+        return result;
+    }
+
+private:
+    static bool compare(const int a, const int b) {
+        return a > b;
+    }
+};

LC-1561/LC-1561.js

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+/**
+ * @param {number[]} piles
+ * @return {number}
+ */
+var maxCoins = function (piles) {
+  /**
+   * Concepts: Sorting
+   */
+
+  let pilesLen = piles.length
+  if (pilesLen < 3) return 0
+
+  piles.sort((a, b) => b - a)
+
+  let result = 0
+  for (let i = 1; i < (2 * pilesLen) / 3; i += 2) {
+    result += piles[i]
+  }
+  return result
+}

LC-1561/LC-1561.ts

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+/**
+ * Concepts: Sorting
+ */
+
+function maxCoins(piles: number[]): number {
+  let pilesLen: number = piles.length
+  if (pilesLen < 3) return 0
+
+  piles.sort((a, b) => b - a)
+
+  let result: number = 0
+  for (let i: number = 1; i < (2 * pilesLen) / 3; i += 2) {
+    result += piles[i]
+  }
+  return result
+}

LC-1561/README.md

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+# LC-1561 - Maximum Number of Coins You Can Get
+
+There are `3n` piles of coins of varying size, you and your friends will take piles of coins as follows:
+
+* In each step, you will choose **any** 3 piles of coins (not necessarily consecutive).
+* Of your choice, Alice will pick the pile with the maximum number of coins.
+* You will pick the next pile with maximum number of coins.
+* Your friend Bob will pick the last pile.
+* Repeat until there are no more piles of coins.
+* Given an array of integers `piles` where `piles[i]` is the number of coins in the `i`-th pile.
+
+Return the maximum number of coins which you can have.
+
+> * Difficulty: **MEDIUM**
+
+---
+## Examples
+
+```
+Input: piles = [2,4,1,2,7,8]
+Output: 9
+Explanation: 
+Choose the triplet (2, 7, 8), Alice Pick the pile with 8 coins, you the pile with 7 coins and Bob the last one.
+Choose the triplet (1, 2, 4), Alice Pick the pile with 4 coins, you the pile with 2 coins and Bob the last one.
+The maximum number of coins which you can have are: 7 + 2 = 9.
+On the other hand if we choose this arrangement (1, 2, 8), (2, 4, 7) you only get 2 + 4 = 6 coins which is not optimal.
+```
+
+```
+Input: piles = [2,4,5]
+Output: 4
+```
+
+```
+Input: piles = [9,8,7,6,5,1,2,3,4]
+Output: 18
+```
+
+---
+## Notes
+
+* `3 <= piles.length <= 10^5`
+* `piles.length % 3 == 0`
+* `1 <= piles[i] <= 10^4`
+
+---
+## Solutions
+
+1. Sorting
+    * Time complexity: $O(n\log{n})$
+    * Space complexity: $O(1)$