547-number-of-provinces.py

class UnionFind:
    def __init__(self, n):
        self.parent = [i for i in range(n)]
        self.rank = [0 for _ in range(n)]
        self.count = n

    def find(self, p):
        while p != self.parent[p]:
            self.parent[p] = self.parent[self.parent[p]]
            p = self.parent[p]
        return p

    def union(self, p, q):
        root_p = self.find(p)
        root_q = self.find(q)
        if root_p == root_q:
            return
        if self.rank[root_p] > self.rank[root_q]:
            self.parent[root_q] = root_p
        elif self.rank[root_p] < self.rank[root_q]:
            self.parent[root_p] = root_q
        else:
            self.parent[root_p] = root_q
            self.rank[root_q] += 1
        self.count -= 1

    def get_count(self):
        return self.count

class Solution:
    def findCircleNum(self, isConnected: List[List[int]]) -> int:
        n = len(isConnected)
        uf = UnionFind(n)
        for p in range(n):
            for q in range(n):
                if isConnected[p][q]:
                    uf.union(p, q)
        return uf.get_count()
    
# time O(n**2), using path compression and union by rank can reduce cost to near O(1)
# space O(n)
# using graph and union find