Description
https://leetcode.com/problems/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]
is1
or0
.isConnected[i][i] == 1
isConnected[i][j] == isConnected[j][i]
Explanation
Conduct depth-first search to find number of provinces.
Python Solution
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).