<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:
        neighbours = [(0, 1), (1, 0), (0, -1), (-1, 0)]
        all_neighbours = [(0, 1), (1, 1), (1, 0), (1, -1), (0, -1), (-1, -1), (-1, 0), (-1, 1)]

        def is_connected(grid, start, removed, neighbours, land_tiles):
            visited = set()
            q = [start]
            while len(q) > 0:
                (r, c) = q.pop()

                if r < 0 or c < 0 or r >= len(grid) or c >= len(grid[0]): continue
                if grid[r][c] == 0: continue
                if grid[r][c] in removed: continue
                visited.add((r, c))
                
                for nr, nc in neighbours:
                    if (r + nr, c + nc) not in visited and (r + nr, c + nc) not in removed:
                        q.append((r + nr, c + nc))
            
            return len(visited) == land_tiles - len(removed)
        
        land_tiles = 0
        shore_tiles = []
        for r in range(len(grid)):
            for c in range(len(grid[0])):
                if grid[r][c] == 1:
                    land_tiles += 1
                    for n in all_neighbours:
                        if n[0] < 0 or n[1] < 0 or n[0] >= len(grid) or n[1] >= len(grid[0]) or grid[n[0]][n[1]] == 0:
                            shore_tiles.append((r, c))
                            break

        if land_tiles == 0 or not is_connected(grid, shore_tiles[0], set(), neighbours, land_tiles):
            return 0
        if land_tiles == 1: return 1
        if land_tiles == 2: return 2
        
        for i in range(len(shore_tiles)):
            start = 1 if i == 0 else 0
            if not is_connected(grid, shore_tiles[start], { shore_tiles[i] }, neighbours, land_tiles):
                return 1

        return 2
1398