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. Examples:

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 : Yes No Yes 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 : Yes No

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

Java

// A Java program to implement Special Social Network
// using Disjoint Set Data Structure.
import java.io.*;
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;
        makeSet();
    }

    // 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)
            return;

        // 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 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))
            System.out.println("Yes");
        else
            System.out.println("No");

        // Check if 1 is a friend of 0
        if (dus.areConnected(0, 1))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

Output:

Yes
No