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
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[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][i] == 1
isConnected[i][j] == isConnected[j][i]
Conduct depth-first search to find number of provinces.
class Solution: def findCircleNum(self, isConnected: List[List[int]]) -> int: count = 0 visited = set() for i in range(len(isConnected)): if i not in visited: self.dfs_helper(isConnected, i, visited) count += 1 return count def dfs_helper(self, isConnected, vertice, visited): visited.add(vertice) for j in range(len(isConnected[vertice])): if isConnected[vertice][j] == 0: continue if j not in visited: self.dfs_helper(isConnected, j, visited)
- Time Complexity: O(N).
- Space Complexity: O(N).