<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>

from queue import SimpleQueue

class Solution:
    def minDays(self, grid: List[List[int]]) -> int:
        SimpleQueue()
        n, m = len(grid), len(grid[0])
        in_bounds = lambda x, y: x >= 0 and x < n and y >= 0 and y < m
        graph = {}
        
        def build_graph(x: int, y: int):
            u = (x, y)
            if u in graph:
                return
            graph[u] = []
            for dx, dy in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
                v = (x + dx, y + dy)
                if in_bounds(*v) and grid[x + dx][y + dy]:
                    graph[u].append(v)
                    build_graph(*v)
        
        for i in range(n):
            for j in range(m):
                if grid[i][j] and (i, j) not in graph:
                    build_graph(i, j)
        
        def count_connected_without(forbidden: Tuple[int, int]) -> int:
            start = None
            for u in graph.keys():
                if u == forbidden:
                    continue
                start = u
                break
            
            visited = {start}
            queue = SimpleQueue()
            queue.put(start)
            while not queue.empty():
                u = queue.get()
                if u in graph:
                    for v in graph[u]:
                        if v == forbidden:
                            continue
                        if v not in visited:
                            queue.put(v)
                            visited.add(v)
            print(start, visited)
            return len(visited)

        if len(graph) == 0 or count_connected_without((-1, -1)) != len(graph):
            return 0
        if len(graph) == 1:
            return 1
        for u in graph.keys():
            c = count_connected_without(u)
            if c < len(graph) - 1:
                return 1
        return 2
        

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