<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>

from collections import deque
class Solution:
    def minDays(self, grid: List[List[int]]) -> int:
        def count_islands(grid):
            def BFS(grid,vis,row,col,n,m):
                delrow=[-1,0,+1,0]
                delcol=[0,+1,0,-1]
                vis[row][col]=True
                q=deque()
                q.append((row,col))
                while q:
                    row,col=q.popleft()
                    for i in range(4):
                        nrow=row+delrow[i]
                        ncol=col+delcol[i]
                        if nrow>=0 and nrow<n and ncol>=0 and ncol<m and not vis[nrow][ncol] and grid[nrow][ncol]==1:
                            vis[nrow][ncol]=True
                            q.append((nrow,ncol))

            n=len(grid)
            m=len(grid[0])
            vis=[[False for j in range(m)] for i in range(n)]
            num_islands=0
            for i in range(n):
                for j in range(m):
                    if grid[i][j]==1 and not vis[i][j]:
                        BFS(grid,vis,i,j,n,m)
                        num_islands+=1

            return num_islands

        n=len(grid)
        m=len(grid[0])
        if count_islands(grid)!=1:
            return 0
        for i in range(n):
            for j in range(m):
                if grid[i][j]==1:
                    grid[i][j]=0
                    if count_islands(grid)!=1:
                        return 1
                    grid[i][j]=1

        return 2


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