547. Number of Provinces

Number of Provinces - LeetCode

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


  • code
class Solution:
    def findCircleNum(self, isConnected: List[List[int]]) -> int:
        n = len(isConnected)
        visited = [False] * n
        
        def dfs(city):
            visited[city] = True
            for nei, connected in enumerate(isConnected[city]):
                if connected == 1 and not visited[nei]:
                    dfs(nei)
                
        province = 0
        for i in range(n):
            if not visited[i]:
                dfs(i)
                province += 1
                
        return province      
  • code
class Solution:
    def findCircleNum(self, isConnected: List[List[int]]) -> int:
        self.isolated = n = len(isConnected)
        root = list(range(n))
        rank = [1] * n
        
        def find(i):
            if i != root[i]:
                root[i] = find(root[i])
            return root[i]
            
        def union(i, j):
            rti = find(i)
            rtj = find(j)
            if rti != rtj:
                if rank[rti] > rank[rtj]:
                    root[rtj] = rti
                elif rank[rti] < rank[rtj]:
                    root[rti] = rtj
                else:
                    root[rti] = rtj
                    rank[rtj] += 1
                self.isolated -= 1 # a merge made
                
                
        for i in range(n):
            for j in range(i+1, n):
                if isConnected[i][j]:
                    union(i, j)
                    
        return self.isolated