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In this post, you will find the solution for Java BitSet-HackerRank Problem. We are providing the correct and tested solutions of coding problems present on HackerRank. If you are not able to solve any problem, then you can take help from our Blog/website.
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Link for the Problem – Java BitSet- Hacker Rank Solution
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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