<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:
    def minDays(self, grid: List[List[int]]) -> int:
        n, m = len(grid), len(grid[0])

        def bfs(grid):
            stack = []
            closed_islands = False
            visited = set()
            for ii, jj in product(range(n), range(m)):
                if grid[ii][jj] and (ii, jj) not in visited:
                    if closed_islands:
                        return True
                    closed_islands = True
                    stack.append((ii, jj))
                    visited.add((ii, jj))
                    while stack:
                        i, j = stack.pop()
                        for x, y in (i-1, j), (i+1, j), (i, j-1), (i, j+1):
                            if 0 <= x < n and 0 <= y < m and grid[x][y] and (x, y) not in visited:
                                stack.append((x, y))
                                visited.add((x, y))
            return not closed_islands

        if bfs(grid):
            return 0
        for i, j in product(range(n), range(m)):
            if grid[i][j]:
                grid[i][j] = 0
                if bfs(grid):
                    return 1
                grid[i][j] = 1
        return 2
        
906