Friend Circle Queries


Problem Statement :


The population of HackerWorld is 10 ^ 9. Initially, none of the people are friends with each other. In order to start a friendship, two persons a and b  have to shake hands, where 1 <= a, b <= 10^9. The friendship relation is transitive, that is if a and b shake hands with each other, a and friends of a become friends with  b and friends of b.

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.

For example, your list of queries is:


Function Description

Complete 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 friendship
Input Format

The first line contains an integer, q, the number of queries to process.
Each of the next q lines consists of two space-separated integers denoting the 2-D array querries.



Output Format

Return an integer array of size q, whose value at index i is the size of largest group present after ith processing the  query.


Solution :



title-img 




                        Solution in C++ :

In  C  ++  :








#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("%d\n",-*(sz.begin()));
	}
	return 0;
}
                    


                        Solution in Java :

In    Java : 




public class Solution {
  public static class DisjointSet {
    private static Map<Integer,Node> map = new HashMap<>();
    private static int max = 0; 
    static class Node {
      long data;
      Node parent;
      int rank;
    }
    public static void makeSet(int data) {
      Node node = new Node();
      node.data = data;
      node.parent = node;
      node.rank = 1;
      map.put(data, node);
    }
    public static boolean union(int data1, int data2) {
        Node node1 = map.get(data1);
        Node node2 = map.get(data2);
        Node parent1 = findSet(node1);
        Node parent2 = findSet(node2);
        if (parent1.data == parent2.data) {
            return false;
        }
        if (parent1.rank >= parent2.rank) {
            parent1.rank = parent1.rank + parent2.rank;
            parent2.parent = parent1;
            max = Math.max(max,parent1.rank);
        } else {
            parent2.rank = parent2.rank+parent1.rank;
            parent1.parent = parent2;
            max = Math.max(max,parent2.rank); 
        }
        return true;
    }
    public static long findSet(int data) {
        return findSet(map.get(data)).data;
    }
    private static Node findSet(Node node) {
        Node parent = node.parent;
        if (parent == node) {
            return parent;
        }
        node.parent = findSet(node.parent);
        return node.parent;
    }
    
}

    static int[] maxCircle(int[][] queries) {
      DisjointSet ds = new DisjointSet();
      int[] arr = new int[queries.length]; 
      for(int[] a : queries) {
        ds.makeSet(a[0]);
        ds.makeSet(a[1]); 
      }

      int i = 0;
      for(int[] a : queries) {
        ds.union(a[0], a[1]); 
        arr[i] = ds.max;
        i++;
      }
      return arr;
    }
                    


                        Solution in Python : 
                            
In   Python3  :







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


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