<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:
        m, n = len(grid), len(grid[0])
        def is_connected(grid: List[List[int]]) -> bool:
            num_islands = 0
            for i, row in enumerate(grid):
                for j, cell in enumerate(row):
                    if cell == 1:
                        stack = [(i - 1, j), (i + 1, j), (i, j - 1), (i, j + 1)]
                        while stack:
                            i, j = stack.pop()
                            if i in range(m) and j in range(n):
                                if grid[i][j] == 1:
                                    stack.append((i - 1, j))
                                    stack.append((i + 1, j))
                                    stack.append((i, j - 1))
                                    stack.append((i, j + 1))
                                    grid[i][j] = -1
                        num_islands += 1
            
            for i in range(m):
                for j in range(n):
                    if grid[i][j]:
                        grid[i][j] = 1
            return num_islands == 1
        
        if not is_connected(grid):
            return 0
        
        for i, row in enumerate(grid):
            for j, cell in enumerate(row):
                if cell == 1:
                    grid[i][j] = 0
                    if is_connected(grid):
                        grid[i][j] = 1
                    else:
                        return 1
        return 2
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