<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:
        if self.countIslands(grid) != 1:
            return 0
        for r in range(len(grid)):
            for c in range(len(grid[0])):
                if grid[r][c] == 1:
                    grid[r][c] = 0
                    if self.countIslands(grid) != 1:
                        return 1
                    grid[r][c] = 1
        return 2
    
    def countIslands(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        unvisited = [[bool(x) for x in row] for row in grid]
        def dfs(r: int, c: int) -> None:
            nonlocal unvisited, m, n
            unvisited[r][c] = False
            if r > 0 and unvisited[r-1][c]:
                dfs(r-1, c)
            if r < m - 1 and unvisited[r+1][c]:
                dfs(r+1, c)
            if c > 0 and unvisited[r][c-1]:
                dfs(r, c-1)
            if c < n - 1 and unvisited[r][c+1]:
                dfs(r, c+1)
        numIslands = 0
        for r in range(m):
            for c in range(n):
                if unvisited[r][c]:
                    numIslands += 1
                    dfs(r, c)
        return numIslands
371