<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:
    vector<pair<int,int>>direction{{1,0},{-1,0},{0,1},{0,-1}};
    vector<pair<int,int>>ans;
    int minDays(vector<vector<int>>& grid) {
        
        int rows=grid.size();
        int cols=grid[0].size();

        vector<vector<int>>discoveryTime(rows,vector<int>(cols));
        vector<vector<int>>lowTime(rows,vector<int>(cols));
        vector<vector<int>>visited(rows,vector<int>(cols));
        vector<vector<int>>parent(rows,vector<int>(cols,-1));
        int landCells=0,landCount=0;
        int timer=0;
        for(int i=0;i<rows;i++)
        {
            for(int j=0;j<cols;j++)
            {
                if(grid[i][j]==1)
                {
                    landCells++;

                    if(!visited[i][j])
                    {
                        findArticulation(discoveryTime,lowTime,grid,visited,i,j,parent,timer);
                        ++landCount;
                    }
                }
            }
        }
        
        if(landCount==0||landCount>=2)
        {
            return 0;
        }
        if(landCells==1)
        return 1;

        if(ans.size()!=0)
        return 1;

        return 2;
    }


    void findArticulation(vector<vector<int>>&discoveryTime,vector<vector<int>>&lowTime,vector<vector<int>>&grid,vector<vector<int>>&visited,int i,int j,vector<vector<int>>&parent,int &timer)
    {
        
        int children=0;
       
        visited[i][j]=1;

        discoveryTime[i][j]=lowTime[i][j]=(++timer);

        for(auto [x,y]:direction)
        {
            if(i+x>=0&&i+x<grid.size()&&j+y>=0&&j+y<grid[0].size()&&grid[i+x][j+y]==1)
            {
                int u=i+x;
                int v=j+y;
                if(!visited[u][v])
                {
                    children++;
                    parent[u][v]=i*j+j;
                    findArticulation(discoveryTime,lowTime,grid,visited,u,v,parent,timer);

                    lowTime[i][j]=min(lowTime[i][j],lowTime[u][v]);

                    if(discoveryTime[i][j]<=lowTime[u][v]&&parent[i][j]!=-1)
                    {
                        ans.push_back({i,j});
                    }
                }
                else if(u*j+v!=parent[i][j])
                {
                    lowTime[i][j]=min(lowTime[i][j],discoveryTime[u][v]);
                }
            }
        }

        if(parent[i][j]==-1&&children>1)
        {
            ans.push_back({i,j});
        }
        
    }
};
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