<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 dfs( int i , int j , vector<vector<int>>&grid , vector<vector<int>>&vis ){
        vis[i][j] = 1;
        
        int dx[] = {-1,0,1,0};
        int dy[] = {0,-1,0,1};

        for(int k = 0 ; k < 4 ; k++){
            int newx = i+dx[k];
            int newy = j+dy[k];
            
            if( newx >= 0 && newy >= 0 && newx < grid.size() && newy < grid[0].size() && !vis[newx][newy] && grid[newx][newy] == 1){
                dfs(newx , newy , grid ,vis);      
            }
        }
    }

    int findcnt(vector<vector<int>>&grid){
        int n = 0;
        vector<vector<int>>vis(30 , vector<int>(30 , 0));
        for(int i = 0 ; i < grid.size() ; i++){
            for( int j = 0 ; j < grid[0].size() ; j++ ){
                if( grid[i][j] == 1 && !vis[i][j]){
                    dfs(i , j ,  grid , vis);
                    n++;
                }
            }
        }
        return n;
    }
    
    int minDays(vector<vector<int>>& grid) {

        int n = findcnt(grid);
        if( n == 0 || n > 1 )return 0;
    
        int ans = 1;
        // guranteed that is there is only one island

        for(int i = 0 ; i < grid.size() ; i++){
            for( int j = 0 ; j < grid[0].size() ; j++ ){
                if( grid[i][j] == 1 ){
                    grid[i][j] = 0;
                    int islands = findcnt(grid);
                    if( islands > 1 || islands == 0 ) return 1;
                    grid[i][j] = 1;
                }
            }
        }

        return 2;
    }
};
39020