<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:
        dirs = [
            (0,1),
            (1,0),
            (-1,0),
            (0,-1),
        ]
        def dfs(y, x, s):
            if (y,x) in s or y < 0 or y >= len(grid) or x < 0 or x >= len(grid[0]) or not grid[y][x]:
                return
            
            s.add((y,x))

            for dy, dx in dirs:
                dfs(y+dy, x+dx, s)
        def check_possible(y,x):
            d = [-1,0,1]
            zero = 0
            for dy in d:
                for dx in d:
                    if dy == dx == 0:
                        continue
                    ny = y + dy
                    nx = x + dx
                    if not (0<=ny<len(grid) and 0<=nx<len(grid[0]) and grid[ny][nx]):
                        zero += 1
                        if zero > 1:
                            return True
            return False
                    
        ones = []
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j]:
                    ones.append((i,j))
        if not ones:
            return 0
        s = set()
        dfs(ones[0][0], ones[0][1], s)
        if len(s) != len(ones):
            return 0
        if len(ones) <= 2:
            return len(ones)
        for idx, (y,x) in enumerate(ones):
            if not check_possible(y,x):
                continue
            grid[y][x] = 0
            s = set()
            ny, nx = ones[(idx+1)%len(ones)]
            dfs(ny,nx, s)
            if len(s) != len(ones)-1:
                return 1
            grid[y][x] = 1
        return 2
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