<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:
        # Get number of island - if != 1, return early
        
        # For each `1` cell get number of `1` neighbors
        # If min neightbors == 1, return 1 else return 2

        # Special case: island of size 2 - both lands must be deleted

        m = len(grid)
        n = len(grid[0])
        delta = [[-1, 0], [1, 0], [0, -1], [0, 1]]

        def countIslands():
            nonlocal m, n, delta

            visited = set()
            islands = 0
            for y in range(m):
                for x in range(n):
                    if grid[y][x] == 0:
                        continue
                    if (x, y) in visited:
                        continue

                    visited.add((x, y))
                    queue = deque()
                    queue.append((x, y))
                    while queue:
                        ox, oy = queue.popleft() # old x, old y
                        for dx, dy in delta:
                            nx = ox + dx
                            ny = oy + dy
                            if nx < 0 or nx >= n or ny < 0 or ny >= m:
                                continue
                            if grid[ny][nx] == 0:
                                continue
                            if (nx, ny) in visited:
                                continue
                            visited.add((nx, ny))
                            queue.append((nx, ny))

                    islands += 1
            return islands

        if countIslands() != 1:
            return 0

        for y1 in range(m):
            for x1 in range(n):
                if grid[y1][x1] == 0:
                    continue

                grid[y1][x1] = 0
                if countIslands() != 1:
                    return 1
                grid[y1][x1] = 1

        return 2

863