<p>You are given an <code>m x n</code> binary grid <code>grid</code> where <code>1</code> represents land and <code>0</code> represents water. An <strong>island</strong> is a maximal <strong>4-directionally</strong> (horizontal or vertical) connected group of <code>1</code>&#39;s.</p>

<p>The grid is said to be <strong>connected</strong> if we have <strong>exactly one island</strong>, otherwise is said <strong>disconnected</strong>.</p>

<p>In one day, we are allowed to change <strong>any </strong>single land cell <code>(1)</code> into a water cell <code>(0)</code>.</p>

<p>Return <em>the minimum number of days to disconnect the grid</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/12/24/land1.jpg" style="width: 500px; height: 169px;" />
<pre>
<strong>Input:</strong> grid = [[0,1,1,0],[0,1,1,0],[0,0,0,0]]

<strong>Output:</strong> 2
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/12/24/land2.jpg" style="width: 404px; height: 85px;" />
<pre>
<strong>Input:</strong> grid = [[1,1]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> Grid of full water is also disconnected ([[1,1]] -&gt; [[0,0]]), 0 islands.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 30</code></li>
	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>

#include <vector>
#include <queue>

using namespace std;

class Solution {
public:
    int minDays(vector<vector<int>>& grid) {
        int rows = grid.size();
        int cols = grid[0].size();
        
        // Initial check for disconnection
        if (isDisconnected(grid)) return 0;

        // Check if changing one cell can disconnect the grid
        for (int r = 0; r < rows; ++r) {
            for (int c = 0; c < cols; ++c) {
                if (grid[r][c] == 1) {
                    grid[r][c] = 0;  // Change cell to water
                    if (isDisconnected(grid)) return 1;
                    grid[r][c] = 1;  // Restore cell
                }
            }
        }

        // Check if changing two cells can disconnect the grid
        for (int r1 = 0; r1 < rows; ++r1) {
            for (int c1 = 0; c1 < cols; ++c1) {
                if (grid[r1][c1] == 1) {
                    grid[r1][c1] = 0;  // Change first cell to water
                    for (int r2 = r1; r2 < rows; ++r2) {
                        for (int c2 = (r2 == r1 ? c1 + 1 : 0); c2 < cols; ++c2) {
                            if (grid[r2][c2] == 1) {
                                grid[r2][c2] = 0;  // Change second cell to water
                                if (isDisconnected(grid)) return 2;
                                grid[r2][c2] = 1;  // Restore second cell
                            }
                        }
                    }
                    grid[r1][c1] = 1;  // Restore first cell
                }
            }
        }

        return 3;  // If all checks fail, return 3 as a fallback
    }

private:
    bool isDisconnected(vector<vector<int>>& grid) {
        int rows = grid.size();
        int cols = grid[0].size();
        vector<vector<bool>> visited(rows, vector<bool>(cols, false));
        
        int initialIslandCount = countIslands(grid, visited);
        
        // If the grid has no land, it's disconnected
        if (initialIslandCount == 0) return true;
        
        // If there's more than one island, it's already disconnected
        if (initialIslandCount > 1) return true;
        
        return false;
    }

    int countIslands(vector<vector<int>>& grid, vector<vector<bool>>& visited) {
        int rows = grid.size();
        int cols = grid[0].size();
        int islandCount = 0;

        vector<pair<int, int>> directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

        for (int r = 0; r < rows; ++r) {
            for (int c = 0; c < cols; ++c) {
                if (grid[r][c] == 1 && !visited[r][c]) {
                    // Start BFS to mark all cells in this island
                    queue<pair<int, int>> q;
                    q.push({r, c});
                    visited[r][c] = true;

                    while (!q.empty()) {
                        auto [x, y] = q.front();
                        q.pop();

                        for (auto [dx, dy] : directions) {
                            int nx = x + dx;
                            int ny = y + dy;

                            if (nx >= 0 && nx < rows && ny >= 0 && ny < cols && grid[nx][ny] == 1 && !visited[nx][ny]) {
                                visited[nx][ny] = true;
                                q.push({nx, ny});
                            }
                        }
                    }

                    islandCount++;
                }
            }
        }
        return islandCount;
    }
};

53020