<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 collections import deque
from typing import Callable, List, Optional


NOT_VISITED = 1
VISITED = 2

class Solution:
    directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]
    
    def traverseOneIsland(self, startI: int, startJ: int) -> int: # BFS
        q = deque()
        q.append((startI, startJ))
        self.grid[startI][startJ] = VISITED
        islandArea = 0

        while q:
            i, j = q.popleft()
            islandArea += 1

            for x, y in self.directions:
                nextI = i + x
                nextJ = j + y

                if 0 <= nextI < self.row and 0 <= nextJ < self.col and self.grid[nextI][nextJ] == NOT_VISITED:
                    q.append((nextI, nextJ))
                    self.grid[nextI][nextJ] = VISITED
        
        return islandArea
    
    def clean(self) -> None:
        for i in range(self.row):
            for j in range(self.col):
                if self.grid[i][j] == VISITED:
                    self.grid[i][j] = NOT_VISITED

    def cntIsland(self) -> int:
        islandCnt = 0
        for i in range(self.row):
            for j in range(self.col):
                if self.grid[i][j] == NOT_VISITED:
                    islandCnt += 1
                    if islandCnt > 1:
                        break
                    self.traverseOneIsland(i, j)

        return islandCnt

    def minDays(self, grid: List[List[int]]) -> int:
        self.grid = grid
        self.row = len(grid)
        self.col = len(grid[0])
        islandArea = 0

        islandCnt = 0
        for i in range(self.row):
            for j in range(self.col):
                if grid[i][j] == NOT_VISITED:
                    islandCnt += 1
                    if islandCnt > 1:
                        return 0
                    islandArea = self.traverseOneIsland(i, j)

        if islandArea <= 2:
            return islandArea

        self.clean()

        for i in range(self.row):
            for j in range(self.col):
                if grid[i][j] == NOT_VISITED:
                    self.grid[i][j] = 0

                    if self.cntIsland() >= 2:
                        return 1

                    self.grid[i][j] = 1
                    self.clean()

        return 2
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