Introduction

In this blog post, we will explore how to solve the LeetCode problem "1861" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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. Example 1: Input: boxGrid = [["#",".","#"]] Output: [["."], ["#"], ["#"]] Example 2: Input: boxGrid = [["#",".","","."], ["#","#","","."]] Output: [["#","."], ["#","#"], ["",""], [".","."]] Example 3: Input: boxGrid = [["#","#","",".","","."], ["#","#","#","",".","."], ["#","#","#",".","#","."]] Output: [[".","#","#"], [".","#","#"], ["#","#",""], ["#","","."], ["#",".",""], ["#",".","."]] Constraints: m == boxGrid.length n == boxGrid[i].length 1 <= m, n <= 500 boxGrid[i][j] is either '#', '', or '.'.

Explanation

Here's a breakdown of the solution:

• Rotation: Create a new n x m matrix to represent the rotated box. Populate it by transposing the original matrix and reversing each row. This simulates a 90-degree clockwise rotation.
• Gravity Simulation: Iterate through each column of the rotated matrix. For each column, simulate gravity by moving stones downwards as far as possible until they encounter an obstacle, another stone, or the bottom.

• In-place Stone Movement: Perform the stone movement in place within the rotated matrix to optimize space.

• Runtime Complexity: O(m*n), Storage Complexity: O(m*n)

Code

    def rotateTheBox(boxGrid):
    m = len(boxGrid)
    n = len(boxGrid[0])

    rotated_box = [['' for _ in range(m)] for _ in range(n)]

    # Rotate the box
    for i in range(m):
        for j in range(n):
            rotated_box[j][m - 1 - i] = boxGrid[i][j]

    # Simulate gravity
    for j in range(m):
        for i in range(n - 1, -1, -1):
            if rotated_box[i][j] == '#':
                rotated_box[i][j] = '.'
                k = i + 1
                while k < n and rotated_box[k][j] == '.':
                    k += 1
                rotated_box[k - 1][j] = '#'

    return rotated_box