Number of Provinces

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:
from collections import defaultdict

class Solution:
    def DFS(self, i, graph, v):
        v.add(i)
        for j in graph[i]:
            if j not in v:
                self.DFS(j, graph, v)
    
    def findCircleNum(self, isConnected: List[List[int]]) -> int:
        n = len(isConnected)
        graph = defaultdict(list)
        for i in range(n):
            for j in range(n):
                if i != j and isConnected[i][j] == 1:
                    graph[i].append(j)
                    graph[j].append(i)
        v = set()
        ctr = 0
        for i in range(n):
            if i not in v:
                self.DFS(i, graph, v)
                ctr += 1
        return ctr

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