Introduction

In this blog post, we will explore how to solve the LeetCode problem "1568" 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 grid grid where 1 represents land and 0 represents water. An island is a maximal 4-directionally (horizontal or vertical) connected group of 1's. The grid is said to be connected if we have exactly one island, otherwise is said disconnected. In one day, we are allowed to change any single land cell (1) into a water cell (0). Return the minimum number of days to disconnect the grid. Example 1: Input: grid = [[0,1,1,0],[0,1,1,0],[0,0,0,0]] Output: 2 Explanation: We need at least 2 days to get a disconnected grid. Change land grid[1][1] and grid[0][2] to water and get 2 disconnected island. Example 2: Input: grid = [[1,1]] Output: 2 Explanation: Grid of full water is also disconnected ([[1,1]] -> [[0,0]]), 0 islands. Constraints: m == grid.length n == grid[i].length 1 <= m, n <= 30 grid[i][j] is either 0 or 1.

Explanation

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

• Connectivity Check: Implement a Depth-First Search (DFS) or Breadth-First Search (BFS) to determine if the grid is connected. This helps determine the initial connectivity and to verify if removing a cell disconnects the grid.
• Edge Case Handling: Handle edge cases where the grid is already disconnected (return 0) or contains only one land cell or only two land cells (return the number of land cells). Also, handle the case where the grid contains only one row or one column of land cells.

• Critical Cell Identification: If the grid is connected and has more than one land cell, iterate through each land cell. Temporarily remove the cell and check if the grid becomes disconnected. If removing any single land cell disconnects the grid, return 1. If not, if there is at least two land cells, we can always remove two cells to disconnect the grid.

• Runtime Complexity: O(m*n * (m*n)), Storage Complexity: O(m*n) - where m and n are the dimensions of the grid.

Code

    def minDays(grid: list[list[int]]) -> int:
    """
    Given a 2D grid representing land (1) and water (0), return the minimum
    number of days to disconnect the grid. In one day, you can change a land
    cell to a water cell.

    Args:
        grid: A 2D list of integers representing the grid.

    Returns:
        The minimum number of days to disconnect the grid.
    """

    m, n = len(grid), len(grid[0])

    def is_connected(grid_copy):
        """
        Helper function to check if the grid is connected using DFS.
        """
        num_islands = 0
        visited = [[False] * n for _ in range(m)]

        def dfs(i, j):
            if i < 0 or i >= m or j < 0 or j >= n or visited[i][j] or grid_copy[i][j] == 0:
                return
            visited[i][j] = True
            dfs(i + 1, j)
            dfs(i - 1, j)
            dfs(i, j + 1)
            dfs(i, j - 1)

        for i in range(m):
            for j in range(n):
                if grid_copy[i][j] == 1 and not visited[i][j]:
                    dfs(i, j)
                    num_islands += 1
        return num_islands == 1 and any(1 in row for row in grid_copy)

    def count_land(grid):
        """
        Helper function to count the number of land cells in the grid.
        """
        count = 0
        for row in grid:
            count += row.count(1)
        return count

    land_count = count_land(grid)

    if land_count == 0:
        return 0
    if land_count <= 2:
        return land_count

    if not is_connected(grid):
        return 0

    # Check if there is a critical cell
    for i in range(m):
        for j in range(n):
            if grid[i][j] == 1:
                # Temporarily remove the cell
                temp_grid = [row[:] for row in grid]  # Create a deep copy
                temp_grid[i][j] = 0

                # Check if the grid is disconnected
                if not is_connected(temp_grid):
                    return 1

    return 2