<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:
        num_rows = len(grid)
        num_cols = len(grid[0])

        def get_id(row: int, col: int):
            nonlocal num_cols
            return num_cols * row + col

        graph: Dict[int, Set[int]] = defaultdict(set)
        for row in range(num_rows):
            for col in range(num_cols):
                if grid[row][col] == 0:
                    continue

                a = get_id(row, col)
                if a not in graph:
                    graph[a] = set()

                if row < num_rows - 1 and grid[row + 1][col] == 1:
                    b = get_id(row + 1, col)
                    graph[a].add(b)
                    graph[b].add(a)

                if col < num_cols - 1 and grid[row][col + 1] == 1:
                    b = get_id(row, col + 1)
                    graph[a].add(b)
                    graph[b].add(a)

        num_islands = self.island_count(graph)
        if num_islands != 1:
            return 0

        for node, nbrs in graph.items():
            graph_copy = deepcopy(graph)
            graph_copy.pop(node)

            for nbr in nbrs:
                graph_copy[nbr].discard(node)

            if self.island_count(graph_copy) != 1:
                return 1

        return 2

    def island_count(self, graph: Dict[int, Set[int]]) -> int:
        visited = set()

        def dfs(node: int):
            if node in visited:
                return

            visited.add(node)

            for nbr in graph[node]:
                dfs(nbr)

        islands = 0
        for node in graph:
            if node not in visited:
                dfs(node)
                islands += 1

                if islands == 2:
                    break

        return islands

18000