<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:
        def dfs(r,c,visited):
            if(r<0 or c<0 or r>=m or c>=n or grid[r][c]==0 or (r,c) in visited):
                return 
            visited.add((r,c))
            ng=[[r,c+1],[r,c-1],[r+1,c],[r-1,c]]
            for nr,nc in ng:
                dfs(nr,nc,visited)
        m=len(grid)
        n=len(grid[0])
        if(m==1 and n==1):
            if(grid[0][0]==1):
                return 1
            else:
                return 0

        visited=set()
        res=0
        fg=0
        for i in range(m):
            for j in range(n):
                if(grid[i][j]==1 and (i,j) not in visited):
                    dfs(i,j,visited)
                    res+=1
        print(res)
        if(res!=1):
            return 0

        v=list(visited)
        for i,j in v:
            visited=set()
            cnt=0
            grid[i][j]=0
            for r2 in range(m):
                for c2 in range(n):
                    if(grid[r2][c2]==1 and (r2,c2) not in visited):
                        dfs(r2,c2,visited)
                        cnt+=1
            if(cnt!=1):
                return 1
            grid[i][j]=1
        return 2
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