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
Problem
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(System.in); 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 = sc.next(); int x = sc.nextInt(); int y = sc.nextInt(); switch (operation) { case "AND": bits[x - 1].and(bits[y - 1]); break; case "OR": bits[x - 1].or(bits[y - 1]); break; case "XOR": bits[x - 1].xor(bits[y - 1]); break; case "FLIP": bits[x - 1].flip(y); break; case "SET": bits[x - 1].set(y); break; default: break; } System.out.println(bits[0].cardinality() + " " + bits[1].cardinality()); } } }
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