HackerRank Friend Circle Queries problem solution YASH PAL, 31 July 20246 February 2026 In this HackerRank Friend Circle Queries problem solution, you will be given q queries. After each query, you need to report the size of the largest friend circle (the largest group of friends) formed after considering that query.Function DescriptionComplete the function maxCircle in the editor below. It must return an array of integers representing the size of the maximum circle of friends after each query.maxCircle has the following parameter(s):queries: an array of integer arrays, each with two elements indicating a new friendshipHackerRank Friend Circle Queries problem solution in Python.#!/bin/python3 import math import os import random import re import sys def init_cmp(mp,x,y): if x not in mp: mp[x]=x if y not in mp: mp[y]=y def init_cc(cc,x,y): if x not in cc: cc[x]=1 if y not in cc: cc[y]=1 def get_parent(mp,x): while mp[x]!=x: x=mp[x] return x # Complete the maxCircle function below. def maxCircle(queries): mp = {} cc = {} max_gp = 0 res = [] for q in queries: init_cmp(mp,q[0],q[1]) init_cc(cc,q[0],q[1]) p1 = get_parent(mp,q[0]) p2 = get_parent(mp,q[1]) if p1!=p2: if cc[p1]>cc[p2]: mp[p2]=p1 cc[p1]=cc[p1]+cc[p2] else: mp[p1]=p2 cc[p2]=cc[p1]+cc[p2] max_gp = max(max_gp,max(cc[p1],cc[p2])) res.append(max_gp) return res if __name__ == '__main__': fptr = open(os.environ['OUTPUT_PATH'], 'w') q = int(input()) queries = [] for _ in range(q): queries.append(list(map(int, input().rstrip().split()))) ans = maxCircle(queries) fptr.write('n'.join(map(str, ans))) fptr.write('n') fptr.close()Friend Circle Queries problem solution in Java.import java.io.*; import java.math.*; import java.security.*; import java.text.*; import java.util.*; import java.util.concurrent.*; import java.util.regex.*; public class Solution { static class UnionFind { Map<Integer, Integer> parents; Map<Integer, Integer> sizes; int max; public UnionFind() { parents = new HashMap<>(); sizes = new HashMap<>(); max = 0; } public void union(int v1, int v2) { if (!parents.containsKey(v1)) { parents.put(v1, v1); sizes.put(v1, 1); } if (!parents.containsKey(v2)) { parents.put(v2, v2); sizes.put(v2, 1); } int p1 = find(v1), p2 = find(v2); if (p1 == p2) return; int s1 = sizes.get(p1), s2 = sizes.get(p2); if (s1 < s2) { parents.put(p1, p2); sizes.put(p2, s1 + s2); if (s1 + s2 > max) max = s1 + s2; }else { parents.put(p2, p1); sizes.put(p1, s1 + s2); if (s1 + s2 > max) max = s1 + s2; } } public int find(int v) { while (parents.get(v) != v) { parents.put(v, parents.get(parents.get(v))); v = parents.get(v); } return v; } } // Complete the maxCircle function below. static int[] maxCircle(int[][] queries) { UnionFind uf = new UnionFind(); int[] res = new int[queries.length]; for (int i = 0; i < queries.length; i++) { uf.union(queries[i][0], queries[i][1]); res[i] = uf.max; } return res; } private static final Scanner scanner = new Scanner(System.in); public static void main(String[] args) throws IOException { BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH"))); int q = scanner.nextInt(); scanner.skip("(rn|[nru2028u2029u0085])?"); int[][] queries = new int[q][2]; for (int i = 0; i < q; i++) { String[] queriesRowItems = scanner.nextLine().split(" "); scanner.skip("(rn|[nru2028u2029u0085])?"); for (int j = 0; j < 2; j++) { int queriesItem = Integer.parseInt(queriesRowItems[j]); queries[i][j] = queriesItem; } } int[] ans = maxCircle(queries); for (int i = 0; i < ans.length; i++) { bufferedWriter.write(String.valueOf(ans[i])); if (i != ans.length - 1) { bufferedWriter.write("n"); } } bufferedWriter.newLine(); bufferedWriter.close(); scanner.close(); } }Problem solution in C++ programming.#include<bits/stdc++.h> using namespace std; const int MAX=500005; int a[MAX], s[MAX]; set<int> sz; void init(int n) { for(int i=0;i<n;i++) { a[i]=i; s[i]=1; } } int root(int x) { while(a[x]!=x) { a[x]=a[a[x]]; x=a[x]; } return x; } void join(int x,int y) { int rx=root(x); int ry=root(y); if(rx==ry) return; if(s[rx]>s[ry]) { s[rx]+=s[ry]; a[ry]=rx; sz.insert(-s[rx]); } else { s[ry]+=s[rx]; a[rx]=ry; sz.insert(-s[ry]); } } int main() { int q; vector<pair<int,int> > queries; map<int,int> m; vector<int> aux; scanf("%d",&q); for(int i=0;i<q;i++) { int x,y; scanf("%d%d",&x,&y); queries.push_back(make_pair(x,y)); aux.push_back(x); aux.push_back(y); } sort(aux.begin(),aux.end()); int curr=1; for(int i=0;i<aux.size();i++) { if(i==0||aux[i]!=aux[i-1]) { m[aux[i]]=curr++; } } for(int i=0;i<q;i++) { queries[i].first=m[queries[i].first]; queries[i].second=m[queries[i].second]; } init(curr); for(int i=0;i<q;i++) { join(queries[i].first,queries[i].second); printf("%dn",-*(sz.begin())); } return 0; } coding problems solutions Hackerrank Problems Solutions interview prepration kit HackerRank