<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:

        ROWS, COLS = len(grid), len(grid[0])

        islands = self.countIslands(grid)
        if islands == 0 or islands > 1: return 0

        # if number of islands is 1, 
        # remove each land block by brute force 
        # and check if the number of islands have changed
        for r in range(ROWS):
            for c in range(COLS):
                if grid[r][c] == 1:
                    grid[r][c] = 0

                    islands = self.countIslands(grid)
                    if islands == 0 or islands > 1: return 1

                    grid[r][c] = 1
        
        # now we need to remove 2 land nodes by default
        return 2


    def countIslands(self, grid: List[List[int]]) -> int:
        ROWS, COLS = len(grid), len(grid[0])

        def inBounds(r, c):
            return (0 <= r < ROWS) and (0 <= c < COLS)

        def dfs(r, c):
            if not inBounds(r, c) or (r, c) in visited or grid[r][c] == 0: return
            visited.add( (r, c) )
            for nr, nc in [(r+1, c), (r, c+1), (r-1, c), (r, c-1)]:
                dfs(nr, nc)

        islands = 0
        visited = set()
        for r in range(ROWS):
            for c in range(COLS):
                if grid[r][c] and (r, c) not in visited:
                    dfs(r, c)
                    islands += 1
        
        return islands

"""
000000
011110
011110
000000

remove from diagonal,

000000
010110
001110
000000

1 island got disconnected,
therefore, 
there is no way to form an example where 
it requires more than 2 removals to disconnect an island
there is always going to be atleast 1 cell thats always covered on only 2 directions
"""
1633