Data Structure to Design a special social network

Consider a special social network where people are called connected if one person is connected to other with any number of intermediate connections. For example if a person x is connected with y and y is connected with z, then x is also considered to be connected with z. We are given a set of friend requests as input. We are also given a set of queries where each query has input pair i and j. For each query, we need to tell whether i and j are connected or not.


Input : Connections :
connect(0, 1), connect(1, 2), connect(0, 3), connect(5, 6), connect (0, 7)
areConnected(2, 7)
areConnected(2, 6)
areConnected(1, 7)
Output :
Explanation : Note that 0 is connected to 2 and 0 is also connected to 7. Therefore 2 and 7 are considered as connected.

Input : Connections :
connect(0, 2), connect(4, 2), connect(1, 3)
areConnected(0, 4)
areConnected(0, 1)
Output :

The idea is to use disjoint set data structure. With this data structure, we can solve all queries in O(1) amortized time.





// A Java program to implement Special Social Network
// using Disjoint Set Data Structure.
import java.util.*;
class DisjointUnionSets {
    int[] rank, parent;
    int n;
    // Constructor
    public DisjointUnionSets(int n)
        rank = new int[n];
        parent = new int[n];
        this.n = n;
    // Creates n sets with single item in each
    void makeSet()
        for (int i = 0; i < n; i++) {
            // Initially, all elements are in
            // their own set.
            parent[i] = i;
    // Returns representative of x's set
    int find(int x)
        // Finds the representative of the set
        // that x is an element of
        if (parent[x] != x) {
            // if x is not the parent of itself
            // Then x is not the representative of
            // his set,
            parent[x] = find(parent[x]);
            // so we recursively call Find on its parent
            // and move i's node directly under the
            // representative of this set
        return parent[x];
    // Unites the set that includes x and the set
    // that includes x
    void connect(int x, int y)
        // Find representatives of two sets
        int xRoot = find(x), yRoot = find(y);
        // Elements are in the same set, no need
        // to unite anything.
        if (xRoot == yRoot)
        // If x's rank is less than y's rank
        if (rank[xRoot] < rank[yRoot])
            // Then move x under y so that depth
            // of tree remains less
            parent[xRoot] = yRoot;
        // Else if y's rank is less than x's rank
        else if (rank[yRoot] < rank[xRoot])
            // Then move y under x so that depth of
            // tree remains less
            parent[yRoot] = xRoot;
        else // if ranks are the same
            // Then move y under x (doesn't matter
            // which one goes where)
            parent[yRoot] = xRoot;
            // And increment the the result tree's
            // rank by 1
            rank[xRoot] = rank[xRoot] + 1;
    boolean areConnected(int i, int j)
        return find(i) == find(j);
// Driver code
public class Main {
    public static void main(String[] args)
        // Let there be 5 persons with ids as
        // 0, 1, 2, 3 and 4
        int n = 5;
        DisjointUnionSets dus = new DisjointUnionSets(n);
        // 0 is a friend of 2
        dus.connect(0, 2);
        // 4 is a friend of 2
        dus.connect(4, 2);
        // 3 is a friend of 1
        dus.connect(3, 1);
        // Check if 4 is a friend of 0
        if (dus.areConnected(0, 4))
        // Check if 1 is a friend of 0
        if (dus.areConnected(0, 1))




Attention reader! Don’t stop learning now. Get hold of all the important CS Theory concepts for SDE interviews with the CS Theory Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using or mail your article to See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.