<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:
        rows, cols = len(grid), len(grid[0])
        lands = set()
        for i in range(rows):
            for j in range(cols):
                if grid[i][j]:
                    lands.add((i, j))
                    
        if len(lands) == 0:
            return 0
        
        def dfs(pos, visited, lands):
            if pos in visited or pos not in lands:
                return
            x, y = pos
            visited.add((x, y))
            dfs((x + 1, y), visited, lands)
            dfs((x - 1, y), visited, lands)
            dfs((x, y + 1), visited, lands)
            dfs((x, y - 1), visited, lands)
        
        def count_connected_lands(lands=lands):
            visited = set()
            dfs(random.choice(list(lands)), visited, lands)
            return len(visited)
        
        if count_connected_lands() < len(lands):
            return 0
        
        for pos in lands:
            curr_lands = lands.difference(set([pos]))
            if len(curr_lands) == 0:
                return 1
            count = count_connected_lands(curr_lands)
            if count < len(curr_lands):
                return 1
        return 2
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