<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>

#include <bits/stdc++.h>
using namespace std;
class Solution {
private:
    std::array<int, 4> dx = {-1,1,0,0};
    std::array<int, 4> dy = {0,0,-1,1};
    
    void dfs(int i, int j, int n, int m, vector<vector<int>> &grid, vector<vector<int>> &vis){
        vis[i][j] = 1;
        
        for(int k=0; k<4; k++){
            int newx = i+dx[k];
            int newy = j+dy[k];
            
            if(newx >= 0 and newy >= 0 and newx < n and newy < m and grid[newx][newy] == 1 and vis[newx][newy] == 0){
                dfs(newx,newy,n,m,grid,vis);
            }
        }
    }
     
private:
    int numIslands(vector<vector<int>>& grid) {
        int n = grid.size();
        int m = grid[0].size();
        
        vector<vector<int>> vis(n,vector<int> (m,0)); //no is visited initially
        int noOfIslands = 0;
        
        for(int i=0; i<n; i++){
            for(int j=0; j<m; j++){
                if(vis[i][j] == 0 and grid[i][j] == 1){
                    dfs(i,j,n,m,grid,vis);
                    noOfIslands++;
                }
            }
        }
        
        return noOfIslands;
    }
    

    
private:
    bool solve(int node, int parent, int &timer, vector<int> &low, vector<int> &tin, vector<int> &vis, vector<int> adj[]){
        
        vis[node] = 1;
        low[node] = tin[node] = timer++;
        int child = 0;
        
        for(auto adjNode: adj[node]){
            if(adjNode == parent) continue;
            
            if(!vis[adjNode]){
                if(solve(adjNode,node,timer,low,tin,vis,adj)){
                    return true;
                }
                
                low[node] = min(low[node],low[adjNode]);
                if(low[adjNode] >= tin[node] and parent != -1){
                    return true;
                }
                child++;
            }
            else{
                low[node] = min(low[node],tin[adjNode]);
            }
        }
        
        if(parent == -1 and child > 1){
            return true;
        }
        return false;
    }
    
private:
    
    
    void make_graph(vector<vector<int>>& grid, vector<int> adj[], int &cnt){
        int n = grid.size(); 
        int m = grid[0].size();
        
        for(int i=0; i<n; i++){
            for(int j=0; j<m; j++){
                if(grid[i][j] == 1){
                    cnt++;
                    
                    for(int k=0; k<4; k++){
                        int newx = i+dx[k];
                        int newy = j+dy[k];

                        if(newx >= 0 and newy >= 0 and newx < n and newy < m and grid[newx][newy] == 1){
                            adj[i*m+j].push_back(newx*m+newy);
                        }
                    }
                }
            }
        }
        
    }

public:
    int minDays(vector<vector<int>>& grid) {
        int n = grid.size(); 
        int m = grid[0].size();
        
        
        int totalIslands = numIslands(grid);
        
        //base case
        if(totalIslands > 1 or totalIslands == 0) return 0;
        
        vector<int> adj[n*m];
        int cnt = 0;
        make_graph(grid,adj,cnt);
        
        //base case
        if(cnt == 1){
            return 1;
        }
        
        
        vector<int> vis(n*m,0), tin(n*m,0), low(n*m,0);
        int timer = 0;

        bool gotAP = false;
        for(int i=0; i<n*m; i++){
            if(!vis[i]){
                if(solve(i,-1,timer,low,tin,vis,adj)){
                    gotAP = true;
                    break;
                }
            }
        }
        
        if(gotAP){
            return 1;
        }
        return 2;
        
    }
};
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