<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])
        visited = [[0] * n for _ in range(m)]
        island_set = set()
        def dfs(x, y, marker):
            nonlocal island_set
            nonlocal m
            nonlocal n
            dirs = [(0, 1), (0, -1), (-1, 0), (1, 0)]
            visited[x][y] = marker
            island_set.add((x,y))
            for dx, dy in dirs:
                nx, ny = x + dx, y + dy
                if 0 <= nx < m and 0 <= ny < n and grid[nx][ny] == 1 and visited[nx][ny] != marker:
                    dfs(nx, ny, marker)
        
        count = 0
        marker = 2 
        for i in range(m):
            for j in range(n):
                if not visited[i][j] and grid[i][j] == 1:
                    if count > 0:
                        return 0
                    else:
                        dfs(i, j, marker)
                        count += 1
                        marker += 1

        if len(island_set) <= 2:
            return len(island_set)
        
        original_set = list(island_set) 
        for i, (x, y) in enumerate(original_set):
            grid[x][y] = 0
            s, t = original_set[(i+1) % len(original_set)]
            island_set.clear()
            dfs(s, t, marker)
            marker += 1
            if len(island_set) < len(original_set) - 1:
                return 1
            grid[x][y] = 1
        
        return 2

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