<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 n = grid.size(), m = grid[0].size();
        vector<vector<int>> visited(n, vector<int>(m, 0));

        auto bfs = [&](int row, int col) {
            queue<pair<int,int>> q;
            q.push({row, col});

            visited[row][col] = 1;

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

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

                for(auto [dr, dc] : directions) {
                    int nr = r + dr;
                    int nc = c + dc;

                    if(nr >= 0 && nr < n && nc >= 0 && nc < m && grid[nr][nc] && !visited[nr][nc]) {
                        visited[nr][nc] = 1;
                        q.push({nr, nc});
                    }
                }
            }
        };
        int count = 0;
        for(int i=0; i<n; i++) {
            for(int j=0; j<m; j++) {
                if(grid[i][j] and !visited[i][j]) {
                    bfs(i,j);
                    count++;
                }
            }
        }

        if(count != 1) return 0;

        vector<pair<int, int>> visitedCells;

        for(int i=0; i<n; i++)
            for(int j=0; j<m; j++)
                if(visited[i][j]) 
                    visitedCells.push_back({i, j});

        for(int i=0; i<visitedCells.size(); i++) {
            count=0;
            auto [row, col] = visitedCells[i];
            visited.assign(n, vector<int>(m, 0));

            grid[row][col] = 0;

            for(int k=0; k<n; k++) {
                for(int j=0; j<m; j++) {
                    if(grid[k][j] and !visited[k][j]) {
                        bfs(k,j);
                        count++;
                    }
                }
            }

            if(count!=1)
                return 1;
            grid[row][col] = 1;
        }
        return 2;
    }
};
30620