<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:
        length = len(grid)
        breadth = len(grid[0])
        def numIslands(grid: List[List[str]]) -> int:
            if not grid:
                return 0

            visits = set()
            islands = 0
            

            def bfs(x, y):
                q = collections.deque()
                visits.add((x, y))
                q.append((x, y))
                while(q):
                    row, col = q.pop()
                    directions = [(1, 0), (0, 1), (-1, 0), (0, -1)]
                    for dr, dc in directions:
                        r = row + dr
                        c = col + dc
                        if r in range(length) and c in range(breadth) and grid[r][c] == 1 and (r,c) not in visits:
                            q.append((r, c))
                            visits.add((r, c))


            for i in range(length):
                for j in range(breadth):
                    if grid[i][j] == 1 and (i, j) not in visits:
                        bfs(i, j)
                        islands += 1

            return islands
        
        if not grid:
            return 0

        if numIslands(grid) != 1:
            return 0

        for i in range(length):
            for j in range(breadth):
                if grid[i][j] == 1:
                    grid[i][j] = 0
                    if numIslands(grid) != 1:
                        return 1
                    grid[i][j] = 1
        return 2
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