<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:
    dirs = ((0, 1), (1, 0), (0, -1), (-1, 0))
    m = len(grid)
    n = len(grid[0])

    def dfs(grid: list[list[int]], i: int, j: int, seen: set[tuple[int, int]]):
      seen.add((i, j))
      for dx, dy in dirs:
        x = i + dx
        y = j + dy
        if x < 0 or x == m or y < 0 or y == n:
          continue
        if grid[x][y] == 0 or (x, y) in seen:
          continue
        dfs(grid, x, y, seen)

    def disconnected(grid: list[list[int]]) -> bool:
      islandsCount = 0
      seen = set()
      for i in range(m):
        for j in range(n):
          if grid[i][j] == 0 or (i, j) in seen:
            continue
          if islandsCount > 1:
            return True
          islandsCount += 1
          dfs(grid, i, j, seen)
      return islandsCount != 1

    if disconnected(grid):
      return 0

    # Try to remove 1 land.
    for i in range(m):
      for j in range(n):
        if grid[i][j] == 1:
          grid[i][j] = 0
          if disconnected(grid):
            return 1
          grid[i][j] = 1

    # Remove 2 lands.
    return 2
820