<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:
        hash_grid = {
            (i, j): val
            for i, row in enumerate(grid)
            for j, val in enumerate(row)
        }
        if disconnected(grid):
            return 0
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j] == 0:
                    continue
                if quick_check(hash_grid, i, j):
                    continue
                grid[i][j] = 0
                if disconnected(grid):
                    return 1
                grid[i][j] = 1
        return 2
        

def quick_check(hash_grid, i, j):
    corners = [(i+1, j+1), (i-1, j-1), (i+1, j-1), (i-1, j+1)]
    return sum(hash_grid.get(c, 0) for c in corners) == 4


def disconnected(grid):
    ones = [
        (i, j)
        for i, row in enumerate(grid)
        for j, val in enumerate(row)
        if val == 1
    ]
    if not ones:
        return True
    stack = [ones[0]]
    seen = set()
    while stack:
        node = stack.pop()
        if node in seen:
            continue
        seen.add(node)
        i, j = node
        nbrs = [(i+1, j), (i-1, j), (i, j+1), (i, j-1)]
        for u, v in nbrs:
            if not (0 <= u < len(grid) and 0 <= v < len(grid[0])):
                continue
            if grid[u][v] == 0:
                continue
            stack.append((u, v))
    return len(seen) != len(ones)
414