<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 collections import defaultdict
class Solution:
    def dfs(self, graph, island, node):
        island.discard(node)
        for neighbour in graph[node]:
            if neighbour in island:
                self.dfs(graph, island, neighbour)

    def minDays(self, grid: List[List[int]]) -> int:
        graph = defaultdict(list)
        rows, cols = len(grid), len(grid[0])
        potential_bridges = set()
        min_degree = 7
        island = set()
        for i in range(rows):
            for j in range(cols):
                if grid[i][j] == 0:
                    continue
                island.add((i, j))
                for x, y in [(i, j + 1), (i, j - 1), (i + 1, j), (i - 1, j)]:
                    if 0 > x or 0 > y or x >= rows or y >= cols:
                        continue
                    if grid[x][y] == 0:
                        continue
                    graph[(i, j)] += [(x, y)]
                min_degree = min(min_degree, len(graph[(i, j)]))
                if min_degree == 2 and (graph[(i, j)][0][0] == graph[(i, j)][1][0] or graph[(i, j)][0][1] == graph[(i, j)][1][1]):
                    potential_bridges.add((i, j))
        if len(graph) <= 1:
            return len(graph)
        visited = set([x for x in island])
        self.dfs(graph, visited, visited.pop())
        if len(visited) > 0:
            return 0
        if min_degree == 1 and len(graph) == 2:
            return 2
        if min_degree == 1:
            return 1
        for bridge in potential_bridges:
            visited = set([x for x in island])
            visited.discard(bridge)
            self.dfs(graph, visited, graph[bridge][0])
            if len(visited) > 0:
                return 1
        return min_degree
        
        
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