<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:
    #define n grid.size()
    #define m grid[0].size()
    #define in(y, x) (y >= 0 && y < n && x >= 0 && x < m)
    pair<int,int> dir[4] = {{0,1},{0,-1},{1,0},{-1,0}};
    pair<int, int> bfs(vector<vector<int>>& grid){
        int mac = 0, num = 0;
        vector<vector<int>> v(n, vector<int>(m, 0));
        for(int i = 0; i < n; ++i)
            for(int j = 0; j < m; ++j){
                if(!grid[i][j] || v[i][j]) continue;
                num++;
                int t2 = 0;
                v[i][j] = 1;
                queue<pair<int,int>> q;
                q.push({i, j});
                while(!q.empty()){
                    int len = q.size();
                    while(len--){
                        auto [y, x] = q.front(); q.pop();
                        t2++;
                        for(auto [a, b]: dir){
                            int ny = y+a, nx = x+b;
                            if(in(ny,nx) && !v[ny][nx] && grid[ny][nx]){
                                q.push({ny, nx});
                                v[ny][nx] = 1;
                            }
                        }
                    }
                }
                mac = max(mac, t2);
            }
        return {num, mac};
    }
    int minDays(vector<vector<int>>& grid) {
        auto [a, b] = bfs(grid);
        if(!a || a > 1) return 0;
        else if(b <= 2) return b;
        for(int i = 0; i < n; ++i)
            for(int j = 0; j < m; ++j){
                if(!grid[i][j]) continue;
                grid[i][j] = 0;
                auto [c, d] = bfs(grid);
                if(c > 1) return 1;
                grid[i][j] = 1;
            }
        return 2;
    }
};
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