HackerRank Java BitSet problem solution

In this HackerRank Java BitSet problem in the java programming language you have Given 2 BitSets, B1 and B2, of size N where all bits in both BitSets are initialized to 0, perform a series of M operations. After each operation, print the number of set bits in the respective BitSets as two space-separated integers on a new line.

HackerRank Java BitSet problem solution.

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
        Scanner scan = new Scanner(System.in);
    int n = scan.nextInt();
    int m = scan.nextInt();

    BitSet b1 = new BitSet(n);
    BitSet b2 = new BitSet(n);

    for (int i = 0; i < m; i++) {
      String opcode = scan.next();
      int num1 = scan.nextInt();
      int num2 = scan.nextInt();

      switch(opcode) {
        case "AND":
          if (num1 == 1 && num2 == 2) {
            b1.and(b2);
          } else if (num1 == 2 && num2 == 1) {
            b2.and(b1);
          } else if (num1 == 1 && num2 == 1) {
            b1.and(b1);
          } else if (num1 == 2 && num2 == 2) {
            b2.and(b2);
          }
          break;
        case "OR":
          if (num1 == 1 && num2 == 2) {
            b1.or(b2);
          } else if (num1 == 2 && num2 == 1) {
            b2.or(b1);
          } else if (num1 == 1 && num2 == 1) {
            b1.or(b1);
          } else if (num1 == 2 && num2 == 2) {
            b2.or(b2);
          }
          break;
        case "XOR":
          if (num1 == 1 && num2 == 2) {
            b1.xor(b2);
          } else if (num1 == 2 && num2 == 1) {
            b2.xor(b1);
          } else if (num1 == 1 && num2 == 1) {
            b1.xor(b1);
          } else if (num1 == 2 && num2 == 2) {
            b2.xor(b2);
          }
          break;
        case "FLIP":
          if (num1 == 1) {
            b1.flip(num2);
          } else if (num1 == 2) {
            b2.flip(num2);
          }
          break;
        case "SET":
          if (num1 == 1) {
            b1.set(num2);
          } else if (num1 == 2) {
            b2.set(num2);
          }
          break;
      }
      System.out.println(b1.cardinality() + " " + b2.cardinality());
    }
    }
}

Second solution

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        
        int n = sc.nextInt();
        BitSet[] a = new BitSet[] {null, new BitSet(n), new BitSet(n)};
        int m = sc.nextInt();
        
        for (int k = 0; k < m; k++) {
            String cmd = sc.next();
            int k1 = sc.nextInt();
            int k2 = sc.nextInt();
            if (cmd.equals("AND")) {
                a[k1].and(a[k2]);
            } else if (cmd.equals("OR")) {
                a[k1].or(a[k2]);
            } else if (cmd.equals("XOR")) {
                a[k1].xor(a[k2]);
            } else if (cmd.equals("FLIP")) {
                a[k1].flip(k2);
            } else if (cmd.equals("SET")) {
                a[k1].set(k2);
            }
            System.out.println(a[1].cardinality() + " " + a[2].cardinality());
        }
    }
}

The solution in java8 programming

import java.io.*;
import java.util.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int N = in.nextInt();
        int M = in.nextInt();
        BitSet[] b = new BitSet[]{new BitSet(N), new BitSet(N)};
        for(int i = 0; i < M; i++) {
            String q = in.next();
            int left = in.nextInt() - 1;
            int right = in.nextInt() - 1;
            if(q.equals("AND")) {
                b[left].and(b[right]);
            }
            if(q.equals("OR")) {
                b[left].or(b[right]);
}
            if(q.equals("XOR")) {
                b[left].xor(b[right]);
}   
            if(q.equals("FLIP")) {
                b[left].flip(N - right - 1);
            }
            if(q.equals("SET")) {
                b[left].set(N - right - 1);
            }
            System.out.println(b[0].cardinality() + " " + b[1].cardinality());
        }
    }
}