<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 = len(grid)
        n = len(grid[0])

        state = {"has_articulation_point": False, "time": 0}
        discovery_time = [[-1] * n for _ in range(m)]
        lowest_reachable_time = [[-1] * n for _ in range(m)]
        parent = [[-1] * n for _ in range(m)]

        def find_articulation_point(row: int, col: int):
            discovery_time[row][col] = lowest_reachable_time[row][col] = state["time"]
            state["time"] += 1
            num_children = 0
            for adj_r, adj_c in [(row - 1, col), (row + 1, col), (row, col - 1), (row, col + 1)]:
                if 0 <= adj_r < m and 0 <= adj_c < n and grid[adj_r][adj_c] == 1:
                    if discovery_time[adj_r][adj_c] == -1:
                        num_children += 1
                        parent[adj_r][adj_c] = row * n + col
                        find_articulation_point(adj_r, adj_c)
                        lowest_reachable_time[row][col] = min(
                            lowest_reachable_time[row][col],
                            lowest_reachable_time[adj_r][adj_c]
                        )
                        if parent[row][col] != -1 and discovery_time[row][col] <= lowest_reachable_time[adj_r][adj_c]:
                            state["has_articulation_point"] = True
                    elif adj_r * n + adj_c != parent[row][col]:
                        lowest_reachable_time[row][col] = min(lowest_reachable_time[row][col], discovery_time[adj_r][adj_c])

            if parent[row][col] == -1 and 1 < num_children:
                state["has_articulation_point"] = True

        num_islands = land_size = 0
        for r in range(m):
            for c in range(n):
                if grid[r][c] == 1:
                    land_size += 1
                    if discovery_time[r][c] == -1:
                        num_islands += 1
                        find_articulation_point(r, c)
        
        if num_islands != 1:
            return 0
        if land_size == 1 or state["has_articulation_point"]:
            return 1
        return 2
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