<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;
    vector<pair<int, int>> adj = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
    vector<vector<int>> vis;

    void dfs(int i, int j, vector<vector<int>> &grid) {
        vis[i][j]=1;
        for(auto u: adj) {
            int ni = i+u.first, nj=j+u.second;
            if(ni<n && ni>=0 && nj<m && nj>=0 && !vis[ni][nj] && grid[ni][nj]) dfs(ni, nj, grid);
        }
    }

    int f(vector<vector<int>> &grid) {
        vis.clear();
        vis.resize(n, vector<int> (m));

        int c = 0;
        for(int i=0; i<n; i++) {
            for(int j=0; j<m; j++) {
                if(!vis[i][j] && grid[i][j]==1) dfs(i, j, grid), c++;
            }
        }

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

        int mini=2;

        for(int i=0; i<n; i++) {
            for(int j=0; j<m; j++) {
                if(grid[i][j]==1) {
                    grid[i][j]=0;
                    int ans = f(grid);
                    if(ans==0 || ans>=2) mini=1;
                    grid[i][j]=1;
                }
            }
        }
        return mini;
    }
};
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