<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) {
        pair<int,int> p1,p2;
        int total = 0;
        int m = grid.size();
        int n = grid[0].size();
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j]==1){
                    total++;
                    p2 = {i,j};
                } 
            }
        }

        //cout<<total<<endl;

        if(total==0) return 0;
        if(total==1) return 1;

        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j]==1){
                    p1 = {i,j};
                    int cnt = count(i,j,m,n,grid);
                    cout<<cnt<<endl;
                    if(cnt!=total) return 0;
                    break;
                } 
            }
        }

        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j]==1){
                    grid[i][j] = 0;
                    int cnt;
                    pair<int,int> p = {i,j};
                    if(p1!=p){
                        cnt = count(p1.first,p1.second,m,n,grid);
                    }else{
                        cnt = count(p2.first,p2.second,m,n,grid);
                    }
                    if(cnt != total - 1) return 1;
                    grid[i][j]=1;
                } 
            }
        }

        


        return 2;
    }

    int count(int i,int j,int m,int n,vector<vector<int>> &grid){
        vector<int> dx={1,-1,0,0};
        vector<int> dy={0,0,1,-1};
        int res = 0;
        queue<pair<int,int>> q;
        q.push({i,j});
        vector<vector<bool>> vis(m,vector<bool>(n,false));
        vis[i][j]= true;
        while(!q.empty()){
            int x = q.front().first;
            int y = q.front().second;
            q.pop();
            res++;
            for(int k=0;k<4;k++){
                int a = x + dx[k];
                int b = y + dy[k];
                if(a>=0 && a<m && b>=0 && b<n && grid[a][b]==1 && !vis[a][b]){
                    vis[a][b] = true;
                    q.push({a,b});
                }
            }
        }
        return res;
    }
};
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