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.
Input: isConnected = [[1,1,0],[1,1,0],[0,0,1]] Output: 2
Input: isConnected = [[1,0,0],[0,1,0],[0,0,1]] Output: 3
- 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]
I solved this problem like this.
I sovled this problem using dfs. If dfs method called, plus 1 on provinces.
class Solution: def findCircleNum(self, isConnected: List[List[int]]) -> int: provinces = 0 visited =  * len(isConnected) n = len(isConnected) def dfs(idx): if visited[idx]: return visited[idx] = 1 for i in range(n): if isConnected[idx][i]: dfs(i) for i in range(n): if not visited[i]: dfs(i) provinces += 1 return provinces