<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 nc(i,j,v,p):
            if not(i>=0 and i<len(grid) and j>=0 and j<len(grid[0]) and grid[i][j] and p[(i,j)]==0):
                return 
            if grid[i][j]:
                grid[i][j]=v
                p[(i,j)]=1
                nc(i+1,j,v,p)
                nc(i-1,j,v,p)
                nc(i,j+1,v,p)
                nc(i,j-1,v,p)
            return 
        def process2():
            process=defaultdict(int)
            c=0
            zc=0
            for i in range(0,len(grid)):
                for j in range(0,len(grid[0])):
                    if grid[i][j]==0:
                        zc+=1
                    if c==0 and grid[i][j] and process[(i,j)]==0:
                        nc(i,j,grid[i][j]+1,process)
                        c+=1
                    elif c==1 and grid[i][j] and process[(i,j)]==0:
                        return True
            if zc==len(grid)*len(grid[0]):
                return True
            return False
        if process2():
           return 0
        def setval(i,j,v):
            grid[i][j]=v
        counter=0
        def process(counter):
                if process2():
                    return counter
                for i in range(0,len(grid)):
                    status=0
                    for j in range(0,len(grid[0])):
                        if grid[i][j]==0:
                            continue
                        print(i,j)
                        if grid[i][j]:
                            grid[i][j]=0
                            counter+=1
                            if process2():
                                    return counter
                            counter-=1
                            grid[i][j]=1
                return 2
            
        return process(counter)
1248