<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();
        int m=grid[0].size();
        vector<vector<int>> check(n,vector<int>(m,-1));
        int dr[]={0,1,-1,0};
        int dc[]={-1,0,0,1};
        int island=0;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(check[i][j]==-1 && grid[i][j]==1){
                    queue<pair<int,int>>q;
                    check[i][j]=0;
                    q.push(make_pair(i,j));
                    while(!q.empty()){
                        int x=q.front().first;
                        int y=q.front().second;
                        q.pop();
                        for(int d=0;d<4;d++){
                            int cx=x+dr[d];
                            int cy=y+dc[d];
                            if(cx>=0&&cy>=0&&cx<n&&cy<m&&check[cx][cy]==-1&&grid[cx][cy]==1){
                                check[cx][cy]=0;
                                check[x][y]++;
                                q.push(make_pair(cx,cy));
                            }
                            else if(cx>=0&&cy>=0&&cx<n&&cy<m&&grid[cx][cy]==1){
                                check[x][y]++;
                            }
                        }
                    }
                    island++;
                }
            }
        }
        if(island!=1){
            return 0;
        }
        if(island==1){
            int cnt=0;
            for(int i=0;i<n;i++){
                for(int j=0;j<m;j++){
                    if(grid[i][j]==1){
                        grid[i][j]=0;
                        vector<vector<int>>check(n,vector<int>(m,-1));
        island=0;
                         for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(check[i][j]==-1 && grid[i][j]==1){
                    queue<pair<int,int>>q;
                    check[i][j]=0;
                    q.push(make_pair(i,j));
                    while(!q.empty()){
                        int x=q.front().first;
                        int y=q.front().second;
                        q.pop();
                        for(int d=0;d<4;d++){
                            int cx=x+dr[d];
                            int cy=y+dc[d];
                            if(cx>=0&&cy>=0&&cx<n&&cy<m&&check[cx][cy]==-1&&grid[cx][cy]==1){
                                check[cx][cy]=0;
                                check[x][y]++;
                                q.push(make_pair(cx,cy));
                            }
                            else if(cx>=0&&cy>=0&&cx<n&&cy<m&&grid[cx][cy]==1){
                                check[x][y]++;
                            }
                        }
                    }
                    island++;
                }
            }
        }
        if(island !=1)
                    {
                        return 1;
                    }
        grid[i][j]=1;
                        
                    }
                }
            }
        }
        return 2;
    }
};
41260