<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:


    void bfs(int row,int col,vector<vector<int>> grid,vector<vector<int>> &visited){
        visited[row][col]=1;
        
        queue<pair<int,int>> q;
        q.push({row,col});
        int n=grid.size();
        int m=grid[0].size();
        
        while(!q.empty()){
            int row=q.front().first;
            int col=q.front().second;
            q.pop();
            
            int delrow[]={0,-1,0,1};
            int delcol[]={-1,0,1,0};

            for(int i=0;i<4;i++){
                int nr=row+delrow[i];
                int mc=col+delcol[i];
                if(nr>=0 && nr<n && mc>=0 && mc<m && grid[nr][mc]==1  && !visited[nr][mc]){
                    visited[nr][mc]=1;
                    q.push({nr,mc});
                }
            }
        }
    }
    
    int numIslands(vector<vector<int>>& grid) {
        
        int n=grid.size();
        int m=grid[0].size();
        int count=0;
        vector<vector<int>> visited(n,vector<int>(m,0));
        
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                
                if(!visited[i][j] && grid[i][j]==1){
                    count++;
                    bfs(i,j,grid,visited);
                }
            }
        }
        return count;
    }

    int minDays(vector<vector<int>>& grid) {
        int island=numIslands(grid);
        if(island==0 or island>1) return 0;
        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){
                    grid[i][j]=0;
                    int island1=numIslands(grid);
                    if(island1==0 or island1>1) return 1;
                    grid[i][j]=1;
                }
            }
        }
        return 2;
    }
};
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