<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 valid(self, x, y, grid):
        m, n = len(grid), len(grid[0])
        return 0 <= x < m and 0 <= y < n

    def dfs(self, x, y, grid):
        self.degree[x][y] = 0
        for dx, dy in self.directions:
            nx, ny = x + dx, y + dy
            if self.valid(nx, ny, grid) and grid[nx][ny] == 1:
                self.degree[x][y] += 1
                if self.degree[nx][ny] == -1:
                    self.dfs(nx, ny, grid)
        

    def get_cc(self, grid):
        cc = 0
        m, n = len(grid), len(grid[0])
        for i in range(m):
            for j in range(n):
                if grid[i][j] == 1 and self.degree[i][j] == -1:
                    cc += 1
                    self.dfs(i, j, grid)
        return cc

    def minDays(self, grid):
        m, n = len(grid), len(grid[0])
        self.directions = [(0, 1), (-1, 0), (1, 0), (0, -1)]
        self.degree = [[-1] * 30 for _ in range(30)]

        cc = self.get_cc(grid)
        if cc != 1:
            return 0

        degree2 = []
        seen = set()

        for x in range(m):
            for y in range(n):
                if 0 <= self.degree[x][y] <= 2:
                    if (x, y) not in seen:
                        degree2.append((x, y))
                        seen.add((x, y))
                    for dx, dy in self.directions:
                        nx, ny = x + dx, y + dy
                        if self.valid(nx, ny, grid) and grid[nx][ny] == 1 and (nx, ny) not in seen:
                            degree2.append((nx, ny))
                            seen.add((nx, ny))

        if len(degree2) == 1:
            return 1

        for x, y in degree2:
            grid[x][y] = 0
            self.degree = [[-1 for _ in range(n)] for _ in range(m)]
            this_cc = self.get_cc(grid)
            if this_cc != 1:
                return 1
            grid[x][y] = 1

        return 2
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