<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>

class Solution {
public:
    int minDays(vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        auto countIslands = [&](vector<vector<int>>& grid) -> int 
        {
            vector<vector<int>> directions = {{0,1},{1,0},{0,-1},{-1,0}};
            vector<vector<int>> visited(m, vector<int>(n, 0));
            int islands = 0;
            auto bfs = [&](int x, int y) 
            {
                queue<pair<int, int>> q;
                q.push({x, y});
                visited[x][y] = 1;
                while (!q.empty()) {
                    auto [cx, cy] = q.front();
                    q.pop();
                    for (auto& dir : directions) 
                    {
                        int nx = cx + dir[0];
                        int ny = cy + dir[1];
                        if (nx >= 0 && nx < m && ny >= 0 && ny < n && grid[nx][ny] == 1 && !visited[nx][ny]) 
                        {
                            visited[nx][ny] = 1;
                            q.push({nx, ny});
                        }
                    }
                }
            };
            
            for (int i = 0; i < m; ++i) 
            {
                for (int j = 0; j < n; ++j) 
                {
                    if (grid[i][j] == 1 && !visited[i][j]) 
                    {
                        ++islands;
                        if (islands > 1) return islands;
                        bfs(i, j);
                    }
                }
            }
            return islands;
        };
        int initialIslands = countIslands(grid);
        if (initialIslands != 1) return 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) 
            {
                if (grid[i][j] == 1) 
                {
                    grid[i][j] = 0;
                    if (countIslands(grid) != 1) 
                    {
                        return 1;
                    }
                    grid[i][j] = 1;
                }
            }
        }
        return 2;
    }
};
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