Java BitSet Hacker Rank Solution

Hello coders, In this post, you will learn how to solve Java BitSet Hacker Rank Solution. This problem is a part of the Java programming series. 

One more thing to add, don’t straight away look for the solutions, first try to solve the problems by yourself. If you find any difficulty after trying several times, then look for the solutions.

Java BitSet- Hacker Rank Solution
Java BitSet Hacker Rank Solution

Java BitSet Hacker Rank Solution


Java’s BitSet class implements a vector of bit values (i.e.:  () or  ()) that grows as needed, allowing us to easily manipulate bits while optimizing space (when compared to other collections). Any element having a bit value of  is called a set bit.

Given  BitSets,  and , of size  where all bits in both BitSets are initialized to , perform a series of  operations. After each operation, print the number of set bits in the respective BitSets as two space-separated integers on a new line.

Problem Statement: Click Here.

Java BitSet Hacker Rank Solution

import java.util.*;
class Solution {
    public static void main(String[] args) {
        Scanner sc = new Scanner(;
        int n = sc.nextInt();
        int m = sc.nextInt();
        BitSet[] bits = {new BitSet(n), new BitSet(n)};
        for (int i = 0; i < m; i++) {
            String operation =;
            int x = sc.nextInt();
            int y = sc.nextInt();
            switch (operation) {
                case "AND":
                    bits[x - 1].and(bits[y - 1]);
                case "OR":
                    bits[x - 1].or(bits[y - 1]);
                case "XOR":
                    bits[x - 1].xor(bits[y - 1]);
                case "FLIP":
                    bits[x - 1].flip(y);
                case "SET":
                    bits[x - 1].set(y);
            System.out.println(bits[0].cardinality() + " " + bits[1].cardinality());

Disclaimer: The above Problem (Java BitSet) is generated by Hackerrank but the Solution is Provided by Chase2Learn. This tutorial is only for Educational and Learning purposes. Authority if any of the queries regarding this post or website fill the following contact form thank you.

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