<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:
        # def is_connected(grid):
        #     visited = [[False]*n for _ in range(m)]
        #     def dfs(x, y):
        #         visited[x][y] = True
        #         for dx, dy in ((0, 1), (0, -1), (1, 0), (-1, 0)):
        #             nx, ny = x + dx, y + dy
        #             if 0 <= nx < m and 0 <= ny < n and not visited[nx][ny] and grid[nx][ny] == 1:
        #                 dfs(nx, ny)
            
        #     found = False
        #     for i in range(m):
        #         for j in range(n):
        #             if grid[i][j] == 1 and not visited[i][j]:
        #                 if found:
        #                     return False  # found another island
        #                 found = True
        #                 dfs(i, j)
        #     return found
        
        def count_islands(grid):
            count = 0
            visited = [[False]*n for _ in range(m)]
            def dfs(x, y):
                visited[x][y] = True
                for dx, dy in ((0, 1), (0, -1), (1, 0), (-1, 0)):
                    nx, ny = x + dx, y + dy
                    if 0 <= nx < m and 0 <= ny < n and not visited[nx][ny] and grid[nx][ny] == 1:
                        dfs(nx, ny)
            
            for i in range(m):
                for j in range(n):
                    if grid[i][j] == 1 and not visited[i][j]:
                        count += 1
                        dfs(i, j)
            return count
        
        m, n = len(grid), len(grid[0])
        
        # If grid is already disconnected
        if count_islands(grid) != 1:
            return 0
        
        # Try to disconnect by removing one cell
        for i in range(m):
            for j in range(n):
                if grid[i][j] == 1:
                    grid[i][j] = 0
                    if count_islands(grid) != 1:
                        return 1
                    grid[i][j] = 1
        
        # If we need to remove two cells
        return 2
17000