Last updated on July 18th, 2024 at 04:51 am

Here, We see Number of Provinces LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

List of all LeetCode Solution

Topics

Depth-First Search, Union-Find

Companies

Bloomberg, Twosigma

Level of Question

Medium

Number of Provinces LeetCode Solution

Problem Statement

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

1. Number of Provinces LeetCode Solution C++

class Solution {
public:
void soln(vector<int>adj[],int node,vector<bool>&vis){
    vis[node] = 1;
    for(auto i:adj[node]){
        if(!vis[i]){
            soln(adj,i,vis);
        }
    }
}
    int findCircleNum(vector<vector<int>>& isConnected) {
        int n = isConnected.size();
        vector<int>adj[n];
        for(int i = 0;i<n;i++){
            for(int j = 0;j<n;j++){
                if(isConnected[i][j]){
                adj[i].push_back(j);
                adj[j].push_back(i);
                }
            }
        }
        vector<bool>vis(n,0);
        int ans= 0;
        for(int i = 0;i<n;i++){
            if(!vis[i]){
                soln(adj,i,vis);
                ans++;
            }
        }
        return ans;
    }
};

2. Number of Provinces LeetCode Solution Java

class Solution {
    public int findCircleNum(int[][] isConnected) {
        int N = isConnected.length;
        boolean[]visited = new boolean[N];
        int count = 0;
        for(int i = 0; i < N ;i++){
            if(!visited[i]){
                count++;
                dfs(isConnected,i,visited);
            }
        }
      return count;  
    }
    private void dfs(int[][]isConnected,int i,boolean[]visited){
        for(int j = 0 ; j < isConnected[i].length ; j++){
            if(!visited[j] && isConnected[i][j] != 0){
                visited[j] = true;
                dfs(isConnected,j,visited);
            }
        }
    }
    
}

3. Number of Provinces Solution JavaScript

var findCircleNum = function(isConnected) {
    class UnionFind {
        constructor(n) {
            this.graph = [...Array(n)].map((_, i) => i);
            this.groups = n;
        }
        find(id) {
            if(this.graph[id] === id) return id;
            this.graph[id] = this.find(this.graph[id]);
            return this.graph[id];
        }
        union(x, y) {
            const rootX = this.find(x);
            const rootY = this.find(y);
            if(rootX !== rootY) {
                this.graph[rootY] = rootX;
                this.groups--;
            }
        }
    }
    const N = isConnected.length, unionfind = new UnionFind(N);
    for(let r = 0; r < N; r++) {
        for(let c = 0; c < N; c++) {
            if(isConnected[r][c]) unionfind.union(r, c);
        }
    }
    return unionfind.groups;
};

4. Number of Provinces Solution Python

class Solution(object):
    def findCircleNum(self, isConnected):
        if not isConnected: return 0
        s = len(isConnected)
        seen = set()
        def dfs(p):
            for q, adj in enumerate(isConnected[p]):
                if (adj == 1) and (q not in seen):
                    seen.add(q)
                    dfs(q)
        cnt = 0
        for i in range(s):
            if i not in seen: 
                dfs(i)
                cnt += 1
        return cnt