<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:
        r=len(grid);c=len(grid[0])
        def  abc(i1,j1,visited):
            if (i1<0 or j1<0 or i1==r or j1==c) or (i1,j1) in visited or grid[i1][j1]==0:
                return
            visited.add((i1,j1))
            neigbor=([i1+1,j1],[i1,j1+1],[i1-1,j1],[i1,j1-1])
            for a,b in neigbor:
                abc(a,b,visited)
        def island():
            res=0
            visited=set()
            for i in range(r):
                for j in range(c):
                    if grid[i][j] ==1 and (i,j) not in visited:
                        abc(i,j,visited)
                        res+=1
            return res
        island()
        if island()!=1:
            return 0
        for u in range(r):
            for v in range(c):
                if grid[u][v]==0:
                    continue
                grid[u][v]=0
                if island()!=1:
                    return 1
                grid[u][v]=1
        return 2        
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