<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 pad(self, grid):
        m = len(grid[0])
        new_grid = [[0] * (m+2)]
        for x in grid:
            new_grid.append([0] + x + [0])
        new_grid.append([0] * (m+2))
        return new_grid

    def dfs(self, grid):
        n, m = len(grid), len(grid[0])
        visited = [[False] * m for _ in range(n)]
        d = [(-1, 0), (1, 0), (0, -1), (0, 1)]
        cnt = 0
        for i in range(1, n-1):
            for j in range(1, m-1):
                if grid[i][j] == 1 and not visited[i][j]:
                    stk = [(i, j)]
                    cnt += 1
                    while stk:
                        x, y = stk.pop()
                        if  visited[x][y]:
                            continue
                        visited[x][y] = True
                        for dx, dy  in d:
                            if grid[x+dx][y+dy] == 1:
                                stk.append((x+dx, y+dy))
        return cnt

    def minDays(self, grid: List[List[int]]) -> int:
        grid = self.pad(grid)
        if self.dfs(grid) != 1:
            return 0
        n, m = len(grid), len(grid[0])
        for i in range(1, n-1):
            for j in range(1, m-1):
                if grid[i][j] == 1:
                    grid[i][j] = 0
                    if self.dfs(grid) != 1:
                        return 1
                    grid[i][j] = 1
        return 2
713