Introduction

In this blog post, we will explore how to solve the LeetCode problem "2132" 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 binary matrix grid where each cell is either 0 (empty) or 1 (occupied). You are then given stamps of size stampHeight x stampWidth. We want to fit the stamps such that they follow the given restrictions and requirements: Cover all the empty cells. Do not cover any of the occupied cells. We can put as many stamps as we want. Stamps can overlap with each other. Stamps are not allowed to be rotated. Stamps must stay completely inside the grid. Return true if it is possible to fit the stamps while following the given restrictions and requirements. Otherwise, return false. Example 1: Input: grid = [[1,0,0,0],[1,0,0,0],[1,0,0,0],[1,0,0,0],[1,0,0,0]], stampHeight = 4, stampWidth = 3 Output: true Explanation: We have two overlapping stamps (labeled 1 and 2 in the image) that are able to cover all the empty cells. Example 2: Input: grid = [[1,0,0,0],[0,1,0,0],[0,0,1,0],[0,0,0,1]], stampHeight = 2, stampWidth = 2 Output: false Explanation: There is no way to fit the stamps onto all the empty cells without the stamps going outside the grid. Constraints: m == grid.length n == grid[r].length 1 <= m, n <= 105 1 <= m n <= 2 105 grid[r][c] is either 0 or 1. 1 <= stampHeight, stampWidth <= 105

Explanation

Here's a breakdown of the approach, followed by the Python code:

• Prefix Sum for Feasibility: Use a 2D prefix sum to efficiently determine if a stamp of size stampHeight x stampWidth can be placed at any given location in the grid without overlapping occupied cells.

• Marking Covered Cells: If a stamp can be placed, mark the corresponding cells as covered.

• Final Check: Iterate through the grid one last time to ensure all empty cells are covered.

• Time & Space Complexity: O(m*n), both time and space, where m and n are the dimensions of the grid.

Code

    def possibleToStamp(grid, stampHeight, stampWidth):
    m, n = len(grid), len(grid[0])

    # 1. Prefix sum to check if stamp can be placed
    prefix_sum = [[0] * (n + 1) for _ in range(m + 1)]
    for i in range(1, m + 1):
        for j in range(1, n + 1):
            prefix_sum[i][j] = prefix_sum[i - 1][j] + prefix_sum[i][j - 1] - prefix_sum[i - 1][j - 1] + grid[i - 1][j - 1]

    # 2. Mark cells that can be covered by a stamp
    covered = [[0] * n for _ in range(m)]
    for i in range(m - stampHeight + 1):
        for j in range(n - stampWidth + 1):
            sum_region = prefix_sum[i + stampHeight][j + stampWidth] - prefix_sum[i][j + stampWidth] - prefix_sum[i + stampHeight][j] + prefix_sum[i][j]
            if sum_region == 0:  # No occupied cells in this region
                for row in range(i, i + stampHeight):
                    for col in range(j, j + stampWidth):
                        covered[row][col] = 1

    # 3. Final check: are all empty cells covered?
    for i in range(m):
        for j in range(n):
            if grid[i][j] == 0 and covered[i][j] == 0:
                return False

    return True