<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 DSU:
    def __init__(self, n):
        self.par = [i for i in range(n)]
        self.rank = [1] * n
    
    def find(self, n):
        if self.par[n] != n:
            self.par[n] = self.find(self.par[n])
        return self.par[n]
    
    def union(self, x, y):
        px, py = self.find(x), self.find(y)
        if self.rank[px] > self.rank[py]: 
            self.par[py] = px
            self.rank[px] += self.rank[py]
        else:
            self.par[px] = py
            self.rank[py] += self.rank[px]

class Solution:
    def minDays(self, grid: List[List[int]]) -> int:
        def dsu_check() -> bool: 
            dsu = DSU(m * n)
            for i in range(m):
                for j in range(n):
                    this_i = i * n + j
                    if grid[i][j] == 1: # land

                        bottom_i = (i + 1) * n + j
                        if i + 1 < m and grid[i + 1][j] == 1: # bottom is land
                            dsu.union(this_i, bottom_i)

                        right_i = i * n + j + 1
                        if j + 1 < n and grid[i][j + 1] == 1: # right is land
                            dsu.union(this_i, right_i)
            
            island = set()
            for i in range(m):
                for j in range(n):
                    this_i = i * n + j
                    dsu.find(this_i)
                    if grid[i][j] == 1:
                        island.add(dsu.par[this_i])

            
            return len(island) > 1 or len(island) == 0

        m, n = len(grid), len(grid[0])
        dsu = DSU(m * n)
        if dsu_check():
            return 0
        
        for ti in range(m):
            for tj in range(n):
                if grid[ti][tj] == 1: 
                    grid[ti][tj] = 0
                    if dsu_check():
                        return 1
                    grid[ti][tj] = 1
        
        return 2
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