<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:
    typedef pair<int, int> pii;
    int bfs(vector<vector<int>> gr, int r, int c) {
        queue<pii> qt;
        int ct = 0;
        for(int i=0; i<r; ++i) {
            for(int j=0; j<c; ++j) {
                if(gr[i][j] == 1) {
                    qt.push({i, j});
                    while(! qt.empty()) {
                        auto p = qt.front(); qt.pop();
                        int x = p.first, y = p.second;
                        if(gr[x][y] != 1) continue;
                        gr[x][y] = 2;
                        if(x-1>=0 && gr[x-1][y] == 1) qt.push({x-1,y});
                        if(x+1<r && gr[x+1][y] == 1) qt.push({x+1, y});
                        if(y-1>=0 && gr[x][y-1] == 1) qt.push({x, y-1});
                        if(y+1<c && gr[x][y+1] == 1) qt.push({x, y+1});
                    }
                    ct++;
                }
            }
        }

        return ct;
    }

    int minDays(vector<vector<int>>& gr) {
        int r = gr.size(), c = r>0 ? gr[0].size() : 0;
        int ct = bfs(gr, r, c);
        if(ct != 1) return 0;
        for(int i=0; i<r; ++i) {
            for(int j=0; j<c; ++j) {
                if(gr[i][j] == 1) {
                    gr[i][j] = 0;
                    if(bfs(gr, r, c) != 1) return 1;
                    gr[i][j] = 1;
                }
            }
        }

        return 2;
    }
};
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