LeetCode 1568: Minimum Number of Days to Disconnect Island

Explanation:

To solve this problem, we can simulate disconnecting the island by converting each land cell to water one by one and checking if the grid becomes disconnected. We can use a depth-first search (DFS) to check if the grid is connected after each conversion. We need to consider two special cases - when the grid has only one island and when the grid is already disconnected.

1. Count the number of islands in the initial grid.
2. If there is only one island, return 0 because the grid is already disconnected.
3. Iterate through each land cell in the grid.
4. For each land cell, convert it to water and perform a DFS to check if the grid becomes disconnected.
5. If the grid becomes disconnected, return the number of days it took to disconnect.
6. If no disconnection occurs after trying all land cells, return 1 (at least 1 day is needed to disconnect).

Time complexity: O(m * n * m * n), where m is the number of rows and n is the number of columns in the grid. Space complexity: O(m * n) for the recursive DFS stack.

:

class Solution {
    public int minDays(int[][] grid) {
        int islands = countIslands(grid);
        if (islands != 1) {
            return 0;
        }
        
        int minDays = 2;
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == 1) {
                    grid[i][j] = 0;
                    if (countIslands(grid) != 1) {
                        return 1;
                    }
                    grid[i][j] = 1;
                }
            }
        }
        
        return minDays;
    }
    
    private int countIslands(int[][] grid) {
        // Implement DFS to count the number of islands
    }
}