<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:
        rows, cols = len(grid), len(grid[0])

        # Initialize discovery time, low value, parent, and articulation point arrays
        disc_time = [[-1 for _ in range(cols)] for _ in range(rows)]
        low_value = [[-1 for _ in range(cols)] for _ in range(rows)]
        parents = [[(-1, -1) for _ in range(cols)] for _ in range(rows)]
        is_ap = [[False for _ in range(cols)] for _ in range(rows)]
        dirs = [(1, 0), (0, 1), (-1, 0), (0, -1)]

        time = 0
        has_ap = False

        def dfs(i, j):
            if grid[i][j] == 0:
                return
            nonlocal time
            nonlocal has_ap
            disc_time[i][j] = time
            low_value[i][j] = time
            time += 1

            child = 0
            for di, dj in dirs:
                ni, nj = i + di, j + dj
                if not (0 <= ni < rows) or not (0 <= nj < cols):
                    continue
                if grid[ni][nj] != 1:
                    continue

                if disc_time[ni][nj] == -1:  # not visited
                    child += 1
                    parents[ni][nj] = (i, j)
                    dfs(ni, nj)
                    low_value[i][j] = min(low_value[i][j], low_value[ni][nj])

                    if parents[i][j] == (-1, -1) and child > 1:
                        is_ap[i][j] = True
                        has_ap = True

                    if parents[i][j] != (-1, -1) and low_value[ni][nj] >= disc_time[i][j]:
                        is_ap[i][j] = True
                        has_ap = True
                elif (ni, nj) != parents[i][j]:
                    low_value[i][j] = min(low_value[i][j], disc_time[ni][nj])

        sccs = 0
        num_ones = 0
        for i in range(rows):
            for j in range(cols):
                if grid[i][j] == 1:
                    num_ones += 1
                if disc_time[i][j] == -1 and grid[i][j] == 1:
                    dfs(i, j)
                    sccs += 1

        if sccs > 1:
            return 0
        elif has_ap:
            return 1
        else:
            if num_ones == 1:
                return 1
            elif num_ones == 0:
                return 0
        return 2
72