<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 x[4]={0,1,0,-1};
    int y[4]={1,0,-1,0};
    void dfs(int i,int j,vector<vector<int>>&a, vector<vector<int>>&vis ){
        vis[i][j]=1;//vis
        int n=a.size();
        int m=a[0].size();

        for(int k=0;k<4;k++){
            int nrow=i+x[k];
            int ncol=j+y[k];
            if(nrow>=0 && nrow<n && ncol>=0 && ncol<m && a[nrow][ncol]==1 && vis[nrow][ncol]==-1){
                dfs(nrow,ncol,a,vis);
            }
        }
        return ;

    }
    int components(vector<vector<int>> &a){
        int n=a.size();
        int m=a[0].size();
        int cmp=0;
        vector<vector<int>> vis(n,vector<int> (m,-1));
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(vis[i][j]==-1 && a[i][j]==1)
                {
                    cmp++;
                    dfs(i,j,a,vis);
                }
            }
        }
        return cmp;
    }
    int minDays(vector<vector<int>>& a) {
        int n=a.size();
        int m=a[0].size();
        int cmp=components(a);
        // return cmp;
        if(cmp==0 || cmp>=2)
        return 0;
        else if(cmp==1){
            for(int i=0;i<n;i++){
                for(int j=0;j<m;j++){
                    if(a[i][j]==1){
                        a[i][j]=0;
                        if(components(a)>=2 || components(a)==0)
                        return 1;
                        a[i][j]=1;
                    }
                }
            }
            return 2;
        }
        return -1;
    }
};
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