<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:
    int n , m;
    int drow[4] = {-1,0,1,0};
    int dcol[4] = {0 ,-1,0, 1};
    void dfs(int i , int j , vector<vector<int>> &grid , vector<vector<int>> &vis){
        vis[i][j] = 1;
        for(int k = 0 ; k<4 ; k++){
            int nrow = i + drow[k];
            int ncol = j+dcol[k];
            if(nrow>=0 and ncol>=0 and nrow<n and ncol<m and !vis[nrow][ncol] and grid[nrow][ncol]){
                dfs(nrow,ncol,grid,vis);
            }
        }

    }
    int minDays(vector<vector<int>>& grid) {
        n = grid.size();
        m = grid[0].size();
        vector<vector<int>> vis(n,vector<int>(m,0));
        vector<pair<int,int>> indices;
        int count = 0;
        for(int i = 0 ; i<n ; i++){
            for(int j = 0 ; j<m ; j++){
                // if(grid[i][j]) indices.push_back({i,j});
                if(!vis[i][j] and grid[i][j]){
                    count++;
                    dfs(i , j , grid, vis);
                } 
            }
        }
        if(count!=1) return 0;

        for(int i = 0 ; i<n ; i++){
            for(int j = 0; j<m ; j++){
                if(grid[i][j]) indices.push_back({i,j});
            }
        }
        
        for(auto ele:indices){
            grid[ele.first][ele.second] = 0;
            vector<vector<int>> vis(n,vector<int>(m,0));
            int count = 0;
            for(int i = 0 ; i<n ; i++){
                for(int j = 0 ; j<m ; j++){
                    // if(grid[i][j]) indices.push_back({i,j});
                    if(!vis[i][j] and grid[i][j]){
                        count++;
                        dfs(i , j , grid, vis);
                    } 
                }
            }
            if(count!=1) return 1;
            grid[ele.first][ele.second] = 1;
        }

        return 2;
    }
};
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