[Leetcode] Medium : Number of Provinces

Description

There are n cities. Some of them are connected, while some are not. If city a is connected directly with city b, and city b is connected directly with city c, then city a is connected indirectly with city c.

A province is a group of directly or indirectly connected cities and no other cities outside of the group.

You are given an n x n matrix isConnected where isConnected[i][j] = 1 if the ith city and the jth city are directly connected, and isConnected[i][j] = 0 otherwise.

Return the total number of provinces.

 

Example 1:

Input: isConnected = [[1,1,0],[1,1,0],[0,0,1]]
Output: 2

Example 2:

Input: isConnected = [[1,0,0],[0,1,0],[0,0,1]]
Output: 3

 

My solution

class Solution:
    def findCircleNum(self, isConnected: List[List[int]]) -> int:
        # 특정 원소가 속한 집합을 찾기
        def find_parent(parent, x):
            # 루트 노드를 찾을 때 까지 재귀 호출
            if parent[x] != x :
                parent[x] = find_parent(parent, parent[x])
            return parent[x]

        # 두 원소가 속한 집합을 합치기
        def union_parent(parent, a, b):
            a = find_parent(parent, a)
            b = find_parent(parent, b)
            if a < b:
                parent[b] = a
            else:
                parent[a] = b

        v = len(isConnected)        
        parent = [0] * v

        for i in range(v):
            parent[i] = i

        for i, con in enumerate(isConnected):
            for j, k in enumerate(con):
                if (k == 1) and (i != j):
                    union_parent(parent, i, j)
        print(parent)
        
        s = set()
        for i in parent:
            s.add(find_parent(parent, i))

        return len(s)

Other Solution

class DSU:
    def __init__(self, N):
        self.par = list(range(N))
    def find(self, x):
        if self.par[x] != x:
            self.par[x] = self.find(self.par[x])
        return self.par[x]
    def union(self, x, y):
        self.par[self.find(x)] = self.find(y)
        
class Solution:
    def findCircleNum(self, isConnected: List[List[int]]) -> int:
        N = len(isConnected)
        dsu = DSU(N)
        for i in range(1, N):
            for j in range(i):
                if isConnected[i][j] == 1:
                    dsu.union(i, j)
        
        s = set()
        for p in dsu.par:
            s.add(dsu.find(p))
        return len(s)

기존에 짰던 코드가 오류가 계속 나타나 사람들이 풀이한 방법을 이용하였다. 기존 방법의 문제는 par리스트 자체를 return을 해 문제가 발생하였음을 확인할 수 있었고, 각 par리스트 안의 원소들의 parent를 찾아주면 정확한 결과를 낼 수 있다. 

문제 출처 : https://leetcode.com/problems/number-of-provinces/

Number of Provinces - LeetCode