<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])

        # Check if grid is already disconnected (0 or 2+ islands)
        if self.count_islands(grid) != 1:
            return 0

        # Check if grid can be disconnected in 1 day
        # by changing a single land cell at a time and counting islands after change
        for i in range(m):
            for j in range(n):
                if grid[i][j] == 1:
                    grid[i][j] = 0
                    if self.count_islands(grid) != 1:
                        return 1
                    grid[i][j] = 1

        # If grid cannot be disconnected in 1 day, it can be split in 2 days
        # by changing the 2 land cells at the thinnest cross-section
        return 2

    def count_islands(self, grid):
        m, n = len(grid), len(grid[0])
        visited = set()
        count = 0

        def dfs(i, j):
            nonlocal m, n
            if i < 0 or j < 0 or i >= m or j >= n or grid[i][j] == 0 or (i, j) in visited:
                return
            visited.add((i, j))
            dfs(i - 1, j)
            dfs(i + 1, j)
            dfs(i, j - 1)
            dfs(i, j + 1)

        for i in range(m):
            for j in range(n):
                if grid[i][j] == 1 and (i, j) not in visited:
                    dfs(i, j)
                    count += 1
        
        return count
948