<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:
        n, m = len(grid), len(grid[0])

        if m == 1 and n > 2:
            for i in range(n - 2):
                if (grid[i][0], grid[i + 1][0], grid[i + 2][0]) == (1, 0, 1):
                    return 0
            return 1
        if n == 1 and m > 2:
            for j in range(m - 2):
                if (grid[0][j], grid[0][j + 1], grid[0][j + 2]) == (1, 0, 1):
                    return 0
            return 1

        def dfs(g, i, j):
            if not (0 <= i < n and 0 <= j < m):
                return
            if g[i][j] in (0, -1):
                return
            g[i][j] = -1
            dfs(g, i + 1, j)
            dfs(g, i - 1, j)
            dfs(g, i, j + 1)
            dfs(g, i, j - 1)
        
        def is_disconnected(g, inplace=False):
            g = g if inplace else copy.deepcopy(g)
            islands = 0
            for i in range(n):
                for j in range(m):
                    if g[i][j] == 1:
                        islands += 1
                        dfs(g, i, j)
                    if islands > 1:
                        return True
            return islands == 0
        
        transformations = {
            (0, 1, 1, 1): (1, 1),
            (1, 0, 1, 1): (1, 0),
            (1, 1, 0, 1): (0, 1),
            (1, 1, 1, 0): (0, 0),

            (0, 1, 0, 1): (0, 1),
            (0, 1, 0, 1): (1, 1),
            (1, 0, 1, 0): (0, 0),
            (1, 0, 1, 0): (1, 0),
            (1, 1, 0, 0): (0, 0),
            (1, 1, 0, 0): (0, 1),
            (0, 0, 1, 1): (1, 0),
            (0, 0, 1, 1): (1, 1),

            (0, 0, 0, 1): (1, 1),
            (0, 0, 1, 0): (1, 0),
            (0, 1, 0, 0): (0, 1),
            (1, 0, 0, 0): (0, 0),
        }

        if is_disconnected(grid):
            return 0
        
        g = copy.deepcopy(grid)
        for i in range(n - 1):
            for j in range(m - 1):
                pattern = (g[i][j], g[i][j + 1], g[i + 1][j], g[i + 1][j + 1])
                if pattern in transformations:
                    di, dj = transformations[pattern]
                    g[i + di][j + dj] = 0
                    if is_disconnected(g, inplace=False):
                        return 1
                    g[i + di][j + dj] = 1
        return 2

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