Introduction

In this blog post, we will explore how to solve the LeetCode problem "1139" 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

Given a 2D grid of 0s and 1s, return the number of elements in the largest square subgrid that has all 1s on its border, or 0 if such a subgrid doesn't exist in the grid. Example 1: Input: grid = [[1,1,1],[1,0,1],[1,1,1]] Output: 9 Example 2: Input: grid = [[1,1,0,0]] Output: 1 Constraints: 1 <= grid.length <= 100 1 <= grid[0].length <= 100 grid[i][j] is 0 or 1

Explanation

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

• Key Idea: Iterate through all possible square subgrid sizes, starting from the largest possible size and decreasing. For each size, check if any subgrid of that size exists with all 1s on its border. Return the area (size * size) if such a subgrid is found, otherwise continue checking smaller sizes.

• Efficient Border Check: For a given size and starting cell, an efficient way to check the border is to verify the top, bottom, left, and right edges contain only 1's.

• Early Exit: If the grid does not contain '1' return 0 immediately.

• Time & Space Complexity:

  • Time Complexity: O(min(m,n)^3 m n), where m and n are the dimensions of the grid. In the worst case, we iterate through all possible square sizes (up to min(m, n)) and for each size, we potentially iterate through all possible starting positions within the grid (m * n), and each border check takes O(size) time.
  • Space Complexity: O(1). The solution uses constant extra space.

Code

    def largest_1_bordered_square(grid: list[list[int]]) -> int:
    """
    Given a 2D grid of 0s and 1s, return the number of elements in the largest square subgrid
    that has all 1s on its border, or 0 if such a subgrid doesn't exist in the grid.
    """
    rows = len(grid)
    cols = len(grid[0])

    if not any(1 in row for row in grid):
        return 0

    max_side = min(rows, cols)

    for side in range(max_side, 0, -1):
        for row in range(rows - side + 1):
            for col in range(cols - side + 1):
                # Check if the current square has all 1s on its border
                valid = True

                # Check top border
                for k in range(col, col + side):
                    if grid[row][k] == 0:
                        valid = False
                        break
                if not valid:
                    continue

                # Check bottom border
                for k in range(col, col + side):
                    if grid[row + side - 1][k] == 0:
                        valid = False
                        break
                if not valid:
                    continue

                # Check left border
                for k in range(row, row + side):
                    if grid[k][col] == 0:
                        valid = False
                        break
                if not valid:
                    continue

                # Check right border
                for k in range(row, row + side):
                    if grid[k][col + side - 1] == 0:
                        valid = False
                        break
                if not valid:
                    continue

                if valid:
                    return side * side

    return 0