<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>

from typing import List

class Solution:
    def minDays(self, grid: List[List[int]]) -> int:
        def is_connected(g):
            """ Returns True if the grid has exactly one island, False otherwise """
            def dfs(x, y):
                if x < 0 or x >= m or y < 0 or y >= n or g[x][y] != 1:
                    return
                g[x][y] = -1  # Mark the cell as visited
                for dx, dy in directions:
                    dfs(x + dx, y + dy)
            
            count = 0
            for i in range(m):
                for j in range(n):
                    if g[i][j] == 1:
                        if count == 1:  # Already found one island, so return False
                            return False
                        count += 1
                        dfs(i, j)
            return count == 1
        
        def check_one_removal():
            for i in range(m):
                for j in range(n):
                    if grid[i][j] == 1:
                        grid[i][j] = 0
                        if not is_connected([row[:] for row in grid]):
                            return True
                        grid[i][j] = 1
            return False
        
        def check_two_removals():
            for i in range(m):
                for j in range(n):
                    if grid[i][j] == 1:
                        grid[i][j] = 0
                        for ii in range(m):
                            for jj in range(n):
                                if grid[ii][jj] == 1:
                                    grid[ii][jj] = 0
                                    if not is_connected([row[:] for row in grid]):
                                        return True
                                    grid[ii][jj] = 1
                        grid[i][j] = 1
            return False
        
        m, n = len(grid), len(grid[0])
        directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]
        
        # Step 1: Check if the grid is already disconnected
        if not is_connected([row[:] for row in grid]):
            return 0
        
        # Step 2: Check if we can disconnect the grid by removing one land cell
        if check_one_removal():
            return 1
        
        # Step 3: If removing one land cell didn't work, check for two cells
        if check_two_removals():
            return 2
        
        # If neither one nor two cell removals work, return 2 by default
        return 2
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