<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>

import copy
class Solution:
	#this is just a helper function for the no_islands function below
    def no_islands_recur(self, grid, i, j, m, n):
        if grid[i][j]==0:
            return
        grid[i][j]=0
        if i-1>=0:
            self.no_islands_recur(grid, i-1, j, m, n)
        if i+1<m:
            self.no_islands_recur(grid, i+1, j, m, n)
        if j-1>=0:
            self.no_islands_recur(grid, i, j-1, m, n)
        if j+1<n:
            self.no_islands_recur(grid, i, j+1, m, n)
    
	
    #find how many islands the given grid has        
    def no_islands(self, grid):
        ret = 0
        m, n = len(grid), len(grid[0])
        for i in range(m):
            for j in range(n):
                if grid[i][j] == 1:
                    ret += 1
                    self.no_islands_recur(grid, i, j, m, n)
        return ret
    
    
    def minDays(self, grid: List[List[int]]) -> int:
        #if we have 0 or more than 1 islands at day 0, return day 0
        time = 0
        grid_copy = copy.deepcopy(grid)
        n = self.no_islands(grid_copy)
        if n!=1:
            return time
        
		#try to remove any land any see if it works
        time = 1
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                grid_copy = copy.deepcopy(grid)
                grid_copy[i][j] = 0
                n = self.no_islands(grid_copy)
                if n!=1:
                    return time
        
		#well then just return 2
        time = 2
        return time   
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