<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:
        m = len(grid)
        n = len(grid[0])
        
        def dfs(r, c, visit):
            if r < 0 or c < 0 or r >= m or c >= n or (r, c) in visit or grid[r][c] == 0:
                return
            
            visit.add((r,c))
            neighbors = [(r, c+1),(r, c-1),(r+1, c),(r-1, c)]

            for nr, nc in neighbors:
                dfs(nr, nc, visit)
        
        def countIsland() -> int:
            visit = set()
            count = 0
            for r in range(m):
                for c in range(n):
                    if grid[r][c] == 1 and (r, c) not in visit:
                        dfs(r,c,visit)
                        count += 1
            return count
        

        # see if remove count is 0
        if countIsland() != 1:
            return 0

        # see if remove count is 1
        for r in range(m):
            for c in range(n):
                if grid[r][c] == 1:
                    grid[r][c] = 0
                    if countIsland() != 1:
                        return 1
                    grid[r][c] = 1

        # see if remove count is 2
        return 2



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