<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:
        self.num_ones = sum([r.count(1) for r in grid])
        if not self.num_ones: return 0
        if self.num_ones == 1: return 1
        if self.is_disconnected(grid, None): return 0
        self.num_ones -= 1
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j] == 1:
                    grid[i][j] = 0
                    c = None
                    if i+1 < len(grid) and grid[i+1][j] == 1: c = (i+1, j)
                    elif j + 1 < len(grid[0]) and grid[i][j+1] == 1: c = (i, j+1)
                    elif i-1 >= 0 and grid[i-1][j] == 1: c = (i-1, j)
                    elif j-1 >= 0 and grid[i][j-1] == 1: c = (i, j-1)
                    if c:
                        if self.is_disconnected(grid, c): return 1
                    grid[i][j] = 1
        return 2
    def is_disconnected(self, grid, coords):
        if self.num_ones == 0: return True
        if self.num_ones == 1: return False
        visited = set()
        if not coords:
            for si in range(len(grid)):
                for sj in range(len(grid[0])):
                    if grid[si][sj] == 1: i, j = si, sj
        else: i, j = coords
        queue = deque([(i, j)])
        while queue:
            i0, j0 = queue.popleft()
            if i0 + 1 < len(grid) and grid[i0+1][j0] == 1 and (i0+1,j0) not in visited:
                queue.append((i0+1, j0))
                visited.add((i0+1, j0))
            if i0 - 1 >= 0 and grid[i0-1][j0] == 1 and (i0-1,j0) not in visited:
                queue.append((i0-1, j0))
                visited.add((i0-1, j0))
            if j0 + 1 < len(grid[0]) and grid[i0][j0+1] == 1 and (i0,j0+1) not in visited:
                queue.append((i0, j0+1))
                visited.add((i0, j0+1))
            if j0 - 1 >= 0 and grid[i0][j0-1] == 1 and (i0,j0-1) not in visited:
                queue.append((i0, j0-1))
                visited.add((i0, j0-1))
        if len(visited) == self.num_ones: return False
        else: return True
670