<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();
        int isl=0;
        int num=0;
        if(disconnected(grid)) return 0;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                int node=i*m+j;
                if(grid[i][j]==1){
                    grid[i][j]=0;
                    if(disconnected(grid)) return 1;
                    grid[i][j]=1;
                }
            }
        }
        return 2;
    }
    bool disconnected(vector<vector<int>>& grid){
        int n=grid.size();
        int m=grid[0].size();
        int isl=0;
        vector<int>vis(n*m,-1);
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                int node=i*m+j;
                if(vis[node]==-1&&grid[i][j]==1){
                    if(isl==1) return true;
                    isl++;
                    dfs(grid,i,j,vis);
                }
            }
        }
        return isl==0;
    }
    void dfs(vector<vector<int>>&grid,int x,int y,vector<int>&vis){
        int n=grid.size();
        int m=grid[0].size();
        int node=x*m+y;
        if(vis[node]==1) return;
        vis[node]=1;
        if(x>0&&grid[x-1][y]==1) dfs(grid,x-1,y,vis);
        if(x<n-1&&grid[x+1][y]==1) dfs(grid,x+1,y,vis);
        if(y>0&&grid[x][y-1]==1) dfs(grid,x,y-1,vis);
        if(y<m-1&&grid[x][y+1]==1) dfs(grid,x,y+1,vis);
        return;
    }
};
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