pascals-triangle.py

# 118. Pascal's Triangle
# 🟢 Easy
#
# https://leetcode.com/problems/pascals-triangle/
#
# Tags: Array - Dynamic Programming

import timeit
from typing import List


# Create the result matrix and then, starting at the third row, fill the
# values for each row based on the values found on the previous row.
#
# Time complexity: O(n^2) - We visit each element twice, when we create
# the empty matrix and when we calculate its value. The number of
# elements has a quadratic relation to the value of the input n and a
# linear O(n) relation to the output.
# Space complexity; O(n^2) - Same as above. Quadratic relation with the
# value of the input and linear with the size of the output.
#
# Runtime 50 ms Beats 43.34%
# Memory 13.9 MB Beats 66.48%
class CreateUpdate:
    def generate(self, numRows: int) -> List[List[int]]:
        # Create the result matrix prefilled with ones. This solves edge
        # cases when numRows < 2.
        matrix = [[1] * (x + 1) for x in range(numRows)]
        # Starting at the 3rd row, fill in the values for all rows.
        for row_idx in range(2, numRows):
            # Generate row sums avoiding the first and last element.
            for col_idx in range(1, row_idx):
                # pos = top_left + top_right when center-aligned.
                matrix[row_idx][col_idx] = (
                    matrix[row_idx - 1][col_idx - 1]
                    + matrix[row_idx - 1][col_idx]
                )
        return matrix


# Similar idea, and O complexity, but slower because append needs to
# grow the matrix, using O(n) time, each time that it doubles in size.
# Between grow operations, append is O(1).
#
# Runtime 63 ms Beats 12.86%
# Memory 14 MB Beats 17.88%
class Append:
    def generate(self, numRows: int) -> List[List[int]]:
        if numRows == 1:
            return [[1]]
        matrix = [[1], [1, 1]]
        if numRows == 2:
            return matrix
        for row_idx in range(2, numRows):
            matrix.append(
                [1]
                + [
                    matrix[row_idx - 1][col_idx - 1]
                    + matrix[row_idx - 1][col_idx]
                    for col_idx in range(1, row_idx)
                ]
                + [1]
            )
        return matrix


def test():
    executors = [
        CreateUpdate,
        Append,
    ]
    tests = [
        [
            5,
            [
                [1],
                [1, 1],
                [1, 2, 1],
                [1, 3, 3, 1],
                [1, 4, 6, 4, 1],
            ],
        ],
        [1, [[1]]],
        [2, [[1], [1, 1]]],
        [3, [[1], [1, 1], [1, 2, 1]]],
    ]
    for executor in executors:
        start = timeit.default_timer()
        for _ in range(1):
            for col, t in enumerate(tests):
                sol = executor()
                result = sol.generate(t[0])
                exp = t[1]
                assert result == exp, (
                    f"\033[93m» {result} <> {exp}\033[91m for"
                    + f" test {col} using \033[1m{executor.__name__}"
                )
        stop = timeit.default_timer()
        used = str(round(stop - start, 5))
        cols = "{0:20}{1:10}{2:10}"
        res = cols.format(executor.__name__, used, "seconds")
        print(f"\033[92m» {res}\033[0m")


test()