547. Number of Provinces

Problem Link

Description

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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

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Python3

class UnionFind:
    def __init__(self, N):
        self._parent = {}
        self._size = {}
        self.connected = N
 
    def union(self, a, b):
        a, b = self.find(a), self.find(b)
        if a == b:
            return
        if self._size[a] < self._size[b]:
            a, b = b, a
        self._parent[b] = a
        self._size[a] += self._size[b]
        self.connected -= 1
 
    def find(self, x):
        if x not in self._parent:
            self._parent[x] = x
            self._size[x] = 1
        while self._parent[x] != x:
            self._parent[x] = self._parent[self._parent[x]]
            x = self._parent[x]
        return x
 
class Solution:
    def findCircleNum(self, isConnected: List[List[int]]) -> int:
        N = len(isConnected)
        uf = UnionFind(N)
 
        for i in range(N):
            for j in range(N):
                if isConnected[i][j] == 1:
                    uf.union(i, j)
        
        return uf.connected