README_EN.md

547. Number of Provinces

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Description

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]

Solutions

Python3

class Solution:
    def findCircleNum(self, isConnected: List[List[int]]) -> int:
        def dfs(i):
            for j in range(n):
                if not visited[j] and isConnected[i][j] == 1:
                    visited[j] = True
                    dfs(j)

        n = len(isConnected)
        visited = [False] * n
        num = 0
        for i in range(n):
            if not visited[i]:
                dfs(i)
                num += 1
        return num

Java

class Solution {
    public int findCircleNum(int[][] isConnected) {
        int n = isConnected.length;
        boolean[] visited = new boolean[n];
        int num = 0;
        for (int i = 0; i < n; ++i) {
            if (!visited[i]) {
                dfs(isConnected, visited, i, n);
                ++num;
            }
        }
        return num;
    }

    private void dfs(int[][] isConnected, boolean[] visited, int i, int n) {
        for (int j = 0; j < n; ++j) {
            if (!visited[j] && isConnected[i][j] == 1) {
                visited[j] = true;
                dfs(isConnected, visited, j, n);
            }
        }
    }
}

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