<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 {
private:
    int countIslands(vector<vector<int>>& grid) {
        int rows = grid.size();
        int cols = grid[0].size();
        vector<vector<bool>> visited(rows, vector<bool>(cols));
        int ans = 0;
        for(int r = 0 ; r < rows; r++) {
            for(int c = 0 ; c < cols; c++){
                if(grid[r][c] == 1 && !visited[r][c]){
                    bfs(grid, rows, cols, r, c, visited);
                    ans++;
                }
            }
        }
        return ans;
    }    
    vector<vector<int>> dirs = {{0,1}, {1, 0}, {0, -1},{-1,0}};
    void bfs(vector<vector<int>>& grid, int rows, int cols, int r, int c, vector<vector<bool>>& visited){
        queue<pair<int, int>> q({{r,c}});
        visited[r][c] = true;
        while(!q.empty()){
            pair<int, int> cur = q.front();
            q.pop();
            for(int i=0; i<dirs.size();i++) {
                int nr = cur.first + dirs[i][0];
                int nc = cur.second + dirs[i][1];
                if(nr>=0&&nr<rows&&nc>=0&&nc<cols&&grid[nr][nc]==1&&!visited[nr][nc]){
                    q.push({nr, nc});
                    visited[nr][nc] = true;
                }
            }
        }
    }
public:
    int minDays(vector<vector<int>>& grid) {
        int rows = grid.size();
        int cols = grid[0].size();
        int count = countIslands(grid);
        if(count > 1 || count == 0) {
            return 0;
        }
        for(int r = 0; r < rows; r++) {
            for(int c = 0; c < cols; c++) {
                if(grid[r][c] == 1){
                    grid[r][c] = 0;
                    int count = countIslands(grid);
                    if(count > 1 || count == 0) {
                        return 1;
                    }
                    grid[r][c] = 1;
                }
            }
        }
        return 2;
    }
};
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