<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])

        j = deque()
        def d(r, c):
            q = deque()
            for x, y in [ (r + 1, c), (r, c + 1), (r - 1, c), (r, c - 1) ]:
                if 0 <= x < m and 0 <= y < n and grid[x][y] != 0:
                    q.append((x, y))
            return q
        def f(r, c):
            grid[r][c] = 2
            q = d(r, c)
            if len(q) >= 2:
                j.append(((r, c), q))
            for x, y in q:
                if grid[x][y] == 1:
                    f(x, y)

        b = 0
        for r, row in enumerate(grid):
            for c, x in enumerate(row):
                if x == 1:
                    if b: return 0
                    f(r, c)
                    b = len(d(r, c)) + 1
        if b <= 1: return b

        def k(s, t, i):
            q = deque([ s ])
            while q:
                x, y = q.popleft()
                grid[x][y] = i
                for a, b in d(x, y):
                    if (a, b) == t:
                        return False
                    if grid[a][b] != i:
                        grid[a][b] = i
                        q.append((a, b))
            return True

        i = 2
        for (r, c), p in j:
            grid[r][c] = 0
            if k(p[0], p[1], (i := i + 1)) or (len(p) >= 3 and k(p[0], p[2], (i := i + 1))) or (len(p) == 4 and k(p[0], p[3], (i := i + 1))):
                return 1
            grid[r][c] = 2
        return 2

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