<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:
        row_length: int = len(grid)
        col_length: int = len(grid[0])
        temp_grid: List[List[int]] = [row[:] for row in grid]
        min_connect: int = 0
        island_amount: int = 0

        def dfs(row: int, col: int) -> None:
            nonlocal temp_grid

            if row < 0 or row >= row_length or col < 0 or col >= col_length or temp_grid[row][col] == 0:
                return

            temp_grid[row][col] = 0

            dfs(row + 1, col)
            dfs(row - 1, col)
            dfs(row, col + 1)
            dfs(row, col - 1)
        
        def count_island() -> int:
            nonlocal temp_grid, row_length, col_length
            temp_island_amount: int = 0
        
            for row in range(row_length):
                for col in range(col_length):
                    if temp_grid[row][col] == 1:
                        dfs(row, col)
                        temp_island_amount += 1
                    
                    if temp_island_amount >= 2:
                        return 2
            
            return temp_island_amount

        island_amount = count_island()
        
        if island_amount == 0 or island_amount == 2:
            return 0

        for row in range(row_length):
            for col in range(col_length):
                if grid[row][col] == 1: 
                    grid[row][col] = 0
                    temp_grid = [row[:] for row in grid]
                    island_amount = count_island()

                    if island_amount == 0 or island_amount >= 2:
                        return 1

                    grid[row][col] = 1
        
        return 2
        

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