Rotating the Box Python Solution with Tests

Tested Python solution for LeetCode 1861 with 22 pytest cases. Generate a practice environment with lcpy.

LeetCode 1861, Medium. Topics: Array, Two Pointers, Matrix. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 22 parametrized pytest cases, and a playground notebook:

lcpy gen -n 1861   # by problem number
lcpy gen -s rotating_the_box   # by problem name

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Problem

You are given an m x n matrix of characters boxGrid representing a side-view of a box. Each cell of the box is one of the following:

• A stone '#'
• A stationary obstacle '*'
• Empty '.'

The box is rotated 90 degrees clockwise, causing some of the stones to fall due to gravity. Each stone falls down until it lands on an obstacle, another stone, or the bottom of the box. Gravity does not affect the obstacles’ positions, and the inertia from the box’s rotation does not affect the stones’ horizontal positions.

It is guaranteed that each stone in boxGrid rests on an obstacle, another stone, or the bottom of the box.

Return an n x m matrix representing the box after the rotation described above.

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Examples

Input: boxGrid = [['#','.','#']]
Output: [['.'],['#'],['#']]

Input: boxGrid = [['#','.','*','.'],['#','#','*','.']]
Output: [['#','.'],['#','#'],['*','*'],['.','.']]

Input: boxGrid = [['#','#','*','.','*','.'],['#','#','#','*','.','.'],['#','#','#','.','#','.']]
Output: [['.','#','#'],['.','#','#'],['#','#','*'],['#','*','.'],['#','.','*'],['#','.','.']]

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Constraints

• m == boxGrid.length
• n == boxGrid[i].length
• 1 <= m, n <= 500
• boxGrid[i][j] is either '#', '*', or '.'.

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Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class Solution:
    # Time: O(m * n)
    # Space: O(m * n) for the result (O(n) auxiliary)
    def rotate_the_box(self, box_grid: list[list[str]]) -> list[list[str]]:
        m, n = len(box_grid), len(box_grid[0])

        def settle(row: list[str]) -> list[str]:
            out = ["."] * n
            write = n - 1
            for i in range(n - 1, -1, -1):
                cell = row[i]
                if cell == "*":
                    out[i] = "*"
                    write = i - 1
                elif cell == "#":
                    out[write] = "#"
                    write -= 1
            return out

        settled = [settle(row) for row in box_grid]
        return [[settled[m - 1 - j][i] for j in range(m)] for i in range(n)]

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Complexity

Time	Space
O(m * n)	O(m * n) for the result (O(n) auxiliary)

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Tags

NeetCode All.