<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:
        
        n, m = len(grid), len(grid[0])
        
        visited = [[False for j in range(m)] for i in range(n)]
        parent = [[-1 for j in range(m)] for i in range(n)]
        low = [[-1 for j in range(m)] for i in range(n)]
        disc = [[-1 for j in range(m)] for i in range(n)]
        
        islands = 0
        hasArticulationPoint = False
        onesCount = 0

        time = 0

        def check_brides(i, j):
            nonlocal hasArticulationPoint
            nonlocal onesCount
            nonlocal time

            onesCount += 1
            visited[i][j] = True

            disc[i][j] = time
            low[i][j] = time
            time += 1

            children = 0

            for x,y in [(i -1, j), (i + 1, j), (i, j - 1), (i, j + 1)]:
                if x >= 0 and x < n and y >= 0 and y < m and grid[x][y] == 1:
                    if not visited[x][y]:
                        parent[x][y] = (i,j)
                        children += 1
                        check_brides(x,y)
                        low[i][j] = min(low[i][j], low[x][y])
                        
                        if parent[i][j] == -1 and children > 1:
                            hasArticulationPoint = True
                        if parent[i][j] != -1 and low[x][y] >= disc[i][j]:
                            hasArticulationPoint = True

                    elif (x,y) != parent[i][j]:
                        low[i][j] = min(low[i][j], disc[x][y])

        for i in range(n):
            for j in range(m):
                if grid[i][j] == 1 and not visited[i][j]:
                    islands += 1
                    check_brides(i, j)
        

        if islands > 1 or islands == 0:
            return 0

        if onesCount == 1 or (hasArticulationPoint and onesCount > 2):
            return 1
    
        return 2
17600