<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) {
        int n = grid.size();
        int m = grid[0].size();
        vis[i][j]=1;
        int dr[]= {-1, 1, 0, 0};
        int dc[]= {0, 0, -1, 1};
        for(int ind=0; ind<4; ind++) {
            int newx= i+dr[ind];
            int newy= j+dc[ind];
            if(newx>=0 && newy>=0 && newx<n && newy<m && !vis[newx][newy] && grid[newx][newy]==1) {
                dfs(newx, newy, grid, vis);
            }
        }
    }
    int numIslands(vector<vector<int>> grid) {
        int n = grid.size();
        int m = grid[0].size();
        vector<vector<int>> vis(n, vector<int>(m,0));
        int ans = 0;
        for(int i=0; i<n; i++) {
            for(int j=0; j<m; j++){
                if(grid[i][j] && !vis[i][j]) {
                    ans++;
                    dfs(i, j, grid, vis);
                }
            }
        }
        return ans;
    }
    int minDays(vector<vector<int>>& grid) {
        int n = grid.size();
        int m = grid[0].size();
        int cnt = numIslands(grid);
        if(!cnt || cnt>1) return 0;
        for(int i=0; i<n; i++) {
            for(int j=0; j<m; j++) {
                if(grid[i][j]==1) {
                    auto temp= grid;
                    temp[i][j]=0;
                    int cnt = numIslands(temp);
                    if(!cnt || cnt>1) return 1;
                }
            }
        }
        return 2;
    }
};
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