<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:
        g = copy.deepcopy(grid)
        g2 = copy.deepcopy(grid)
        c = 0

        def dfs(i,j,g):
            g[i][j] = 0
            if i > 0 and g[i-1][j] == 1:
                dfs(i-1,j,g)
            if j > 0 and g[i][j-1] == 1:
                dfs(i,j-1,g)
            if i < len(g) - 1 and g[i+1][j] == 1:
                dfs(i+1,j,g)
            if j < len(g[0]) - 1 and g[i][j+1] == 1:
                dfs(i,j+1,g)
            
            return

        for i in range((len(g))):
            for j in range(len(g[0])):
                if g[i][j] == 1:
                  dfs(i,j,g)  
                  c += 1

                  if c == 2:
                    return 0
        if c == 0:
            return 0
        
        c = 0
        
        for i in range(len(g2)):
            for j in range(len(g2[0])):
                if g2[i][j] == 1:
                    g3 = copy.deepcopy(g2)
                    g3[i][j] = 0

                    for m in range(len(g2)):
                        for n in range(len(g2[0])):
                            if g3[m][n] == 1:
                                dfs(m,n,g3)
                                c += 1

                                if c == 2:
                                    return 1
                    if c == 0:
                        return 1
                    
                    c = 0
        
        return 2

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