"""
547. Number of Provinces
Medium

4377

217

Add to List

Share
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]
Accepted
390,889
Submissions
626,420
"""

class Solution:
    def findCircleNum(self, isConnected: List[List[int]]) -> int:
        n = len(isConnected)
        parent = { i:i for i in range(1, n + 1)}

        def find(i):
            p = parent[i]
            if p == i:
                return i
            parent[i] = find(p)
            return parent[i]

        for i in range(n):
            for j in range(n):
                if isConnected[i][j] == 1:
                    parent[find(i + 1)] = find(j + 1)
        provinces =  set()
        for i in range(1, n + 1):
            provinces.add(find(i))

        return len(provinces)
