<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:
        # greedy - just disconnect one grid
        m = len(grid)  # row
        n = len(grid[0])  # col
        def isValid(pos: List[int]) -> bool:
            return pos[0] >= 0 and pos[0] < m and pos[1] >= 0 and pos[1] < n
        def numOfIsland() -> int:
            island = 0
            q = deque()
            visited = [[False] * n for _ in range(m)]
            for i in range(m):
                for j in range(n):
                    if grid[i][j] == 0:
                        visited[i][j] = True
                    elif not visited[i][j]:
                        # bfs
                        q.append((i, j))  # init
                        visited[i][j] = True
                        while len(q) > 0:
                            cell = q.popleft()
                            # neighbour
                            up = (cell[0] - 1, cell[1])
                            down = (cell[0] + 1, cell[1])
                            left = (cell[0], cell[1] - 1)
                            right = (cell[0], cell[1] + 1)
                            if isValid(up) and grid[up[0]][up[1]] == 1 and not visited[up[0]][up[1]]:
                                q.append(up)
                                visited[up[0]][up[1]] = True
                            if isValid(down) and grid[down[0]][down[1]] == 1 and not visited[down[0]][down[1]]:
                                q.append(down)
                                visited[down[0]][down[1]] = True
                            if isValid(left) and grid[left[0]][left[1]] == 1 and not visited[left[0]][left[1]]:
                                q.append(left)
                                visited[left[0]][left[1]] = True
                            if isValid(right) and grid[right[0]][right[1]] == 1 and not visited[right[0]][right[1]]:
                                q.append(right)
                                visited[right[0]][right[1]] = True
                        island += 1
                        if island > 1:
                            return island
            return island
        if numOfIsland() != 1:  # numOfIsland == 0 or 2
            return 0
        for i in range(m):
            for j in range(n):
                if grid[i][j] == 1:
                    grid[i][j] = 0  # temp
                    if numOfIsland() != 1:
                        return 1
                    grid[i][j] = 1  # reset
        return 2
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