<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 code(x, y):
            return 9000000 + x * 1000 + y

        # Build adj
        m, n = len(grid), len(grid[0])
        d = {}
        for x in range(m):
            for y in range(n):
                if grid[x][y] == 0:
                    continue

                c = code(x, y)
                d[c] = []
                for dx, dy in [[0, 1], [0, -1], [1, 0], [-1, 0]]:
                    xx, yy = x + dx, y + dy
                    if 0 <= xx < m and 0 <= yy < n and grid[xx][yy]:
                        d[c].append(code(xx, yy))

        def visit(x, ban):
            visited = set()

            def dfs(v):
                if v != ban and v not in visited:
                    visited.add(v)
                    for x in d[v]:
                        dfs(x)
            
            dfs(x)

            return len(visited)

        lands = len(d)
        if lands <= 1:
            return lands
        elif visit(next(iter(d.keys())), -42) < lands:
            return 0

        for k, items in d.items():
            if len(items) > 1 and visit(items[0], k) < lands - 1:
                return 1

        return min(len(d), 2)

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