<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>

const int sz=901;
#define pb push_back
#define ll long long
class Solution {
public:


    int dx[4]={0,-1,1,0};
    int dy[4]={-1,0,0,1};
    
    set<int>adj[sz];
    int visited[sz];
int low[sz];
int high[sz];
int check[sz];
int timer=0;
void dfs(int node,int par){
    visited[node]=1;
    int child=0;
    low[node]=timer;
    high[node]=timer++;
    for(auto it:adj[node]){
        if(it==par){
            continue;
        }
        else if(visited[it]==1){
            low[node]=min(low[node],high[it]);
        }
        else{
            child++;
            dfs(it,node);
            low[node]=min(low[it],low[node]);
             if (low[it] >= high[node] && par!=-1){
                check[node]=1;
                }
        }
    }
    if(par==-1 && child>1){
        check[node]=1;
    }
}
    int minDays(vector<vector<int>>& grid) {
        map<pair<int,int>,int>mp;
        int total=0;
        int n=grid.size();
        int m=grid[0].size();
        int ind=0;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(grid[i][j]){
                    total++;
                }
                mp[{i,j}]=ind++;
            }
        }
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
            for(int k=0;k<4;k++){
                int nx=i+dx[k];
                int ny=j+dy[k];
                if(nx<n && ny<m && nx>=0 && ny>=0 && grid[i][j]==1 && grid[nx][ny]==1){
                    adj[mp[{i,j}]].insert(mp[{nx,ny}]);
                    adj[mp[{nx,ny}]].insert(mp[{i,j}]);
                }
            }
            }
        }
        int cnt=0;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(visited[mp[{i,j}]]==0 && grid[i][j]==1 ){
                    cnt++;
                    dfs(mp[{i,j}],-1);
                }
            }
        }
        if(total==0){
            return 0;
        }
         if(total==1){
            return 1;
        }
        if(cnt>1){
            return 0;
        }
        for(int i=0;i<sz;i++){
            if(check[i]==1){
                return 1;
            }
        }
       
        return 2;
    }
};
18300