<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:
    
	//below code for finding number of connected components
    int no_of_conn(vector<vector<int>>& grid){
        
        int visit[30][30] = {0};
        int count =0;
        queue<pair<int,int>> q;
        for(int i=0;i<grid.size();i++){
            
            for(int j=0;j<grid[i].size();j++){
                
                if(grid[i][j]==1 && visit[i][j]==0){
                    count++;
                    q.push({i,j});
                    visit[i][j] = 1;
                    
                        while(!q.empty()){
                            
                            int s = q.size();
                            
                            for(int i=0;i<s;i++){
                                auto p = q.front();
                                q.pop();
                                int x = p.first;
                                int y = p.second;
    
                                if(x+1<grid.size() && visit[x+1][y]==0 && grid[x+1][y]==1){
                                    q.push({x+1,y});
                                    visit[x+1][y] = 1;
                                }
                                if(y+1<grid[0].size() && visit[x][y+1]==0 && grid[x][y+1]==1){
                                    q.push({x,y+1});
                                    visit[x][y+1] = 1;
                                }
                                if(x-1>=0 && visit[x-1][y]==0 && grid[x-1][y]==1){
                                    q.push({x-1,y});
                                    visit[x-1][y] = 1;
                                }
                                if(y-1>=0 && visit[x][y-1]==0 && grid[x][y-1]==1){
                                    q.push({x,y-1});
                                    visit[x][y-1] = 1;
                                }
                            }
                        }
                        
                }
                
            }
            
        }
        
        
        return count;
       
    }
    
    int minDays(vector<vector<int>>& grid) {
        
        int num = no_of_conn(grid);
        
        if(num!=1){
            return 0;
        }
        
        
            
            vector<pair<int,int>> ind;
            
            for(int i=0;i<grid.size();i++){
                
                for(int j=0;j<grid[i].size();j++){
                    
                    if(grid[i][j]==1){
                        grid[i][j] = 0;//make that 0
                        
						// check for number of conn components
                        if(no_of_conn(grid)!=1){
                            return 1;
                        }
                        
                        grid[i][j] =1;//make that 1 again as it was not the desired land
                        
                    }
                    
                }
                
            }
            
            
            
        return 2; //at max we need only 2 days to disconnect any island ...Thinnk about it. :)
        
        
    }
};
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