<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:
        directions = [(0,1), (0,-1), (1,0), (-1,0)]
        from collections import deque
        ans, land = self.bfs_land(grid, directions)
        if not ans:
            return 0
        for i, j in land:
            grid[i][j] = 0
            ans = self.bfs_ans(grid, land, i, j, directions)
            if not ans:
                return 1
            grid[i][j] = 1
        return 2

    def bfs_land(self, grid, directions):
        land = []
        visited = set()
        queue = deque()
        island = 0
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j] == 1:
                    land.append((i, j))
                if grid[i][j] == 0 or (i, j) in visited:
                    continue
                queue.append((i, j))
                island += 1
                while queue:
                    r, c = queue.popleft()
                    for r1, c1 in directions:
                        next_r, next_c = r + r1, c +c1
                        if not self.is_valid(next_r, next_c, grid, visited):
                            continue
                        queue.append((next_r, next_c))
                        visited.add((next_r, next_c))       
        if island == 1:
            return True, land
        else:
            return False, land

    def is_valid(self, r, c, grid, visited):
        if not (0 <= r < len(grid) and 0 <= c < len(grid[0])):
            return False
        if grid[r][c] == 0:
            return False
        if (r, c) in visited:
            return False
        return True

    def bfs_ans(self, grid, land, i, j, directions):
        visited = set()
        island = 0
        queue = deque()
        for r, c in land:
            if (r, c) == (i, j):
                continue
            if (r, c) in visited:
                continue
            queue.append((r, c))
            island += 1
            while queue:
                r, c = queue.popleft()
                for r1, c1 in directions:
                    next_r, next_c = r + r1, c + c1
                    if not self.is_valid(next_r, next_c, grid, visited):
                        continue
                    queue.append((next_r, next_c))
                    visited.add((next_r, next_c)) 
        if island == 1:
            return True
        else:
            return False
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