<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, n = len(grid), len(grid[0])
        D = [(1, 0), (-1, 0), (0, 1), (0, -1)]
        total_one = sum(sum(row) for row in grid)

        def check():
            visited = set()
            island_count = 0

            while len(visited) < m * n:
                for r in range(m):
                    for c in range(n):
                        if (r, c) in visited:
                            continue
                        
                        visited.add((r, c))

                        if grid[r][c] == 1:
                            island_count += 1
                            visited.add((r, c))
                            q = deque([(r, c)])
                            
                            while q:
                                cur_r, cur_c = q[0][0], q[0][1]
                                q.popleft()
                                
                                for d in D:
                                    nxt_r, nxt_c = cur_r + d[0], cur_c + d[1]
                                    if nxt_r < 0 or nxt_r >=m or nxt_c < 0 or nxt_c >= n or (nxt_r, nxt_c) in visited:
                                        continue
                                    
                                    visited.add((nxt_r, nxt_c))
                                    if grid[nxt_r][nxt_c] == 1:
                                        q.append((nxt_r, nxt_c))

            return island_count
        
        count = check()
        if count > 1 or count == 0:
            return 0
        elif count == 1 and total_one == 1:
            return 1
        elif total_one == 2:
            return 2
        elif total_one == 3:
            return 1
        
        for r in range(m):
            for c in range(n):
                if grid[r][c] == 1:
                    grid[r][c] = 0
                    
                    if check() > 1:
                        return 1

                    grid[r][c] = 1

        return 2

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