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547. Number of Provinces

Difficulty Topics

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

 

Constraints:

  • 1 <= n <= 200
  • n == isConnected.length
  • n == isConnected[i].length
  • isConnected[i][j] is 1 or 0.
  • isConnected[i][i] == 1
  • isConnected[i][j] == isConnected[j][i]

Solution

number-of-provinces.py
class DSU:
    def __init__(self, n):
        self.graph = list(range(n))

    def find(self, x):
        if self.graph[x] != x:
            self.graph[x] = self.find(self.graph[x])

        return self.graph[x]

    def union(self, x, y):
        ux, uy = self.find(x), self.find(y)
        self.graph[ux] = uy

    def connected(self, x, y):
        return self.find(x) == self.find(y)

class Solution:
    def findCircleNum(self, G: List[List[int]]) -> int:
        n = len(G)
        uf = DSU(n)

        for node in range(n):
            for nei in range(n):
                if node == nei or G[node][nei] == 0: continue

                uf.union(node, nei)

        parents = set()

        for node in range(n):
            p = uf.find(node)
            parents.add(p)

        return len(parents)