<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 UnionFind:
    def __init__(self, n):
        self.father = [i for i in range(n)]

    def union(self, a, b):
        fa = self.find(a)
        fb = self.find(b)
        if fa != fb:
            self.father[fa] = fb

    def find(self, a):
        fa = self.father[a]
        if fa != a:
            self.father[a] = self.find(fa)
        return self.father[a]

class Solution:
    def closedIsland(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        dirs = [(1,0), (-1,0), (0,1), (0,-1)]

        def get_id(i, j):
            return i * n + j
        
        def at_boundary(i, j):
            return i == 0 or j == 0 or i == m - 1 or j == n - 1
        
        res = 0
        uf = UnionFind(m * n)
        for i in range(m):
            for j in range(n):
                if grid[i][j] == 0:
                    c_id = get_id(i, j)
                    for di, dj in dirs:
                        ni, nj = i + di, j + dj
                        if 0 <= ni < m and 0 <= nj < n and grid[ni][nj] == 0:
                            uf.union(c_id, get_id(ni, nj))

        roots = set()
        for i in range(m):
            for j in range(n):
                if grid[i][j] == 0:
                    roots.add(uf.find(get_id(i, j)))
        
        print(roots)
        
        for i in range(m):
            for j in range(n):
                if grid[i][j] == 0 and at_boundary(i, j):
                    to_remove = uf.find(get_id(i, j))
                    if to_remove in roots:
                        roots.remove(to_remove)
        return len(roots)
17400