<p>Given a 2D&nbsp;<code>grid</code> consists of <code>0s</code> (land)&nbsp;and <code>1s</code> (water).&nbsp; An <em>island</em> is a maximal 4-directionally connected group of <code><font face="monospace">0</font>s</code> and a <em>closed island</em>&nbsp;is an island <strong>totally</strong>&nbsp;(all left, top, right, bottom) surrounded by <code>1s.</code></p>

<p>Return the number of <em>closed islands</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2019/10/31/sample_3_1610.png" style="width: 240px; height: 120px;" /></p>

<pre>
<strong>Input:</strong> grid = [[1,1,1,1,1,1,1,0],[1,0,0,0,0,1,1,0],[1,0,1,0,1,1,1,0],[1,0,0,0,0,1,0,1],[1,1,1,1,1,1,1,0]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> 
Islands in gray are closed because they are completely surrounded by water (group of 1s).</pre>

<p><strong class="example">Example 2:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2019/10/31/sample_4_1610.png" style="width: 160px; height: 80px;" /></p>

<pre>
<strong>Input:</strong> grid = [[0,0,1,0,0],[0,1,0,1,0],[0,1,1,1,0]]
<strong>Output:</strong> 1
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> grid = [[1,1,1,1,1,1,1],
&nbsp;              [1,0,0,0,0,0,1],
&nbsp;              [1,0,1,1,1,0,1],
&nbsp;              [1,0,1,0,1,0,1],
&nbsp;              [1,0,1,1,1,0,1],
&nbsp;              [1,0,0,0,0,0,1],
               [1,1,1,1,1,1,1]]
<strong>Output:</strong> 2
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= grid.length, grid[0].length &lt;= 100</code></li>
	<li><code>0 &lt;= grid[i][j] &lt;=1</code></li>
</ul>

class Solution:
    def closedIsland(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        djs = DisjointSet(m, n)
        for i, row in enumerate(grid):
            for j, cell in enumerate(row):
                if cell == 1:
                    continue
                if i in (0, m - 1) or j in (0, n - 1):
                    djs.union(i, j, m - 1, n)
                if i < m - 1 and grid[i + 1][j] == 0:
                    djs.union(i, j, i + 1, j)
                if j < n - 1 and grid[i][j + 1] == 0:
                    djs.union(i, j, i, j + 1)
        return djs.num_closed(grid)
        
        
class DisjointSet:
    
    def __init__(self, m: int, n: int) -> None:
        self.root = [[(i, j) for j in range(n)] for i in range(m)]
        self.rank = [[0] * n for _ in range(m)]
        self.root[-1].append((-1, -1))
        self.rank[-1].append(10**9)
        
    def find(self, i: int, j: int) -> tuple[int, int]:
        if self.root[i][j] != (i, j):
            self.root[i][j] = self.find(*self.root[i][j])
        return self.root[i][j]
    
    def union(self, i: int, j: int, v: int, u: int) -> None:
        ri, rj = self.find(i, j)
        rv, ru = self.find(v, u)
        if (ri, rj) == (rv, ru):
            return
        kij, kvu = self.rank[ri][rj], self.rank[rv][ru]
        if kij < kvu:
            self.root[ri][rj] = (rv, ru)
        else:
            self.root[rv][ru] = (ri, rj)
            if kij == kvu:
                self.rank[ri][rj] += 1
        
    def num_closed(self, grid: list[list[int]]) -> int:
        roots = set()
        m, n = len(grid), len(grid[0])
        for i, row in enumerate(grid):
            for j, cell in enumerate(row):
                if cell == 1:
                    continue
                root = self.find(i, j)
                if root != (-1, -1):
                    roots.add(root)
        return len(roots)
17600