Number of Provinces LeetCode Solution

Problem – 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.

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]

Number of Provinces LeetCode Solution in Java

public class Solution {
    public void dfs(int[][] M, int[] visited, int i) {
        for (int j = 0; j < M.length; j++) {
            if (M[i][j] == 1 && visited[j] == 0) {
                visited[j] = 1;
                dfs(M, visited, j);
            }
        }
    }
    public int findCircleNum(int[][] M) {
        int[] visited = new int[M.length];
        int count = 0;
        for (int i = 0; i < M.length; i++) {
            if (visited[i] == 0) {
                dfs(M, visited, i);
                count++;
            }
        }
        return count;
    }
}

Number of Provinces LeetCode Solution in Python

def findCircleNum(self, A):
    N = len(A)
    seen = set()
    def dfs(node):
        for nei, adj in enumerate(A[node]):
            if adj and nei not in seen:
                seen.add(nei)
                dfs(nei)
    
    ans = 0
    for i in xrange(N):
        if i not in seen:
            dfs(i)
            ans += 1
    return ans

Number of Provinces LeetCode Solution in C++

class Solution {
public:
    int findCircleNum(vector<vector<int>>& M) {
        if (M.empty()) return 0;
        int n = M.size();
        vector<bool> visited(n, false);
        int groups = 0;
        for (int i = 0; i < visited.size(); i++) {
            groups += !visited[i] ? dfs(i, M, visited), 1 : 0;
        }
        return groups;
    }

private:
    void dfs(int i, vector<vector<int>>& M, vector<bool>& visited) {
        visited[i] = true;
        for (int j = 0; j < visited.size(); j++) {
            if (i != j && M[i][j] && !visited[j]) {
                dfs(j, M, visited);
            }
        }
    }
};
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