<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:
    void bfs( int sr , int sc,vector<vector<int>>&grid , vector<vector<int>>&vis ){
        vis[sr][sc] = 1;
        queue<pair<int,int>>q;
        q.push({sr,sc});

        int m = grid.size();
        int n = grid[0].size();

        int delrow[] = {0 , 1 , 0 , -1};
        int delcol[] = {-1 , 0 , 1 , 0};
        while( !q.empty()){
            int row = q.front().first;
            int col = q.front().second;
            q.pop();

            for( int i = 0 ; i < 4 ; i++){
                int nrow = delrow[i] + row;
                int ncol = delcol[i] + col;

                if( nrow >= 0 && nrow < m && ncol >= 0 && ncol < n && grid[nrow][ncol] == 1 && !vis[nrow][ncol]){
                    vis[nrow][ncol] = 1;
                    q.push({nrow,ncol});
                }
            }
        }
    }
    int minDays(vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        vector<vector<int>>vis( m , vector<int>( n , 0));
        int noOfisland = 0;
        for( int i = 0 ; i < m ; i++){
            for( int j = 0 ; j < n ; j++){
                if( grid[i][j] == 1 && !vis[i][j]){
                    bfs( i , j , grid , vis);
                    noOfisland++;
                }
            }
        }
        if( noOfisland > 1 || noOfisland == 0){
            return 0;
        }
        for( int i = 0 ; i < m ; i++){
            for( int j = 0 ; j < n ; j++){
                vector<vector<int>>vis1( m , vector<int>( n , 0));
                if(grid[i][j] == 1 ){
                    grid[i][j] = 0;
                    int count = 0;
                        for( int a = 0 ; a < m ; a++){
                            for( int b = 0 ; b < n ; b++){
                                if( grid[a][b] == 1 && !vis1[a][b]){
                                    bfs( a , b , grid , vis1);
                                    count++;
                                }
                            }
                        }
                    if( count != 1){
                        return 1;
                    }
                    grid[i][j] = 1;
                }
            }
        }
        return 2;
    }
};
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