<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:
        def countIsland(grid):
            def dfs(ri, ci):
                if -1 < ri < R and -1 < ci < C and not seen[ri][ci] and grid[ri][ci] == 1:
                    seen[ri][ci] = True
                    dfs(ri-1, ci)
                    dfs(ri, ci-1)
                    dfs(ri+1, ci)
                    dfs(ri, ci+1)

            seen = [[False] * C for _ in range(R)]
            res = 0
            for ri in range(R):
                for ci in range(C):
                    if grid[ri][ci] == 1 and not seen[ri][ci]:
                        res += 1
                        dfs(ri, ci)
            return res
        
        def countLand(grid):
            res = 0
            for ri in range(R):
                for ci in range(C):
                    res += grid[ri][ci]
            return res
        
        def countMinEdge(grid):
            res = inf
            tedge = [[inf] * C for _ in range(R)]
            for ri in range(R):
                for ci in range(C):
                    if grid[ri][ci] == 1:
                        for tri, tci in [(ri-1, ci), (ri, ci-1), (ri+1, ci), (ri, ci+1)]:
                            if -1 < tri < R and -1 < tci < C and grid[tri][tci] == 1:
                                if tedge[tri][tci] == inf:
                                    tedge[tri][tci] = 0
                                tedge[tri][tci] += 1
            
            for ri in range(R):
                for ci in range(C):
                    res = min(res, tedge[ri][ci])
            return res
        
        def isOneToSplit(grid):
            def isSameGroup(ri1, ci1, ri2, ci2):
                if not (-1 < ri1 < R) or not (-1 < ri2 < R) or not (-1 < ci1 < C) or not (-1 < ci2 < C):
                    return True
                if grid[ri1][ci1] != 1 or grid[ri2][ci2] != 1:
                    return True

                seen = [[False] * C for _ in range(R)] 
                cur = [(ri1, ci1)]
                for ri, ci in cur:
                    if ri == ri2 and ci == ci2:
                        return True
                    for tri, tci in [(ri-1, ci), (ri, ci-1), (ri+1, ci), (ri, ci+1)]:
                        if -1 < tri < R and -1 < tci < C and not seen[tri][tci] and grid[tri][tci] == 1:
                            seen[tri][tci] = True
                            cur.append((tri, tci))
                return False
            
            for ri in range(R):
                for ci in range(C):
                    if grid[ri][ci] == 1:
                        grid[ri][ci] = 0

                        res = not isSameGroup(ri-1, ci, ri+1, ci) or not isSameGroup(ri, ci-1, ri, ci+1) or not isSameGroup(ri-1, ci, ri, ci-1) or not isSameGroup(ri-1, ci, ri, ci+1) or not isSameGroup(ri, ci-1, ri+1, ci) or not isSameGroup(ri+1, ci, ri, ci+1)

                        grid[ri][ci] = 1

                        if res:
                            return True
            return False

        
        R, C = len(grid), len(grid[0])

        island_count = countIsland(grid)
        land_count = countLand(grid)
        min_edge_count = countMinEdge(grid)

        if island_count != 1:
            return 0
        
        if land_count == 1:
            return 1
        
        if min_edge_count == 1 and land_count > 2:
            return 1

        if isOneToSplit(grid):
            return 1
        
        return 2
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