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    import java.io.*;
    import java.util.*;
    
    public class Solution {
    
        public static void main(String[] args) {
            Scanner scanner = new Scanner(System.in);
            
            int n = scanner.nextInt();
            int m = scanner.nextInt();
            
            BitSet[] b = new BitSet[2];
            b[0] = new BitSet(n);
            b[1] = new BitSet(n);
            
            for (int i=0; i<m; i++) {
                String operation = scanner.next();
                int operand1 = scanner.nextInt();
                int operand2 = scanner.nextInt();
                
                if ("AND".equals(operation)) {
                    b[operand1-1].and(b[operand2-1]);
                } else if ("OR".equals(operation)) {
                    b[operand1-1].or(b[operand2-1]);
                } else if ("XOR".equals(operation)) {
                    b[operand1-1].xor(b[operand2-1]);
                } else if ("FLIP".equals(operation)) {
                    b[operand1-1].flip(operand2);
                } else if ("SET".equals(operation)) {
                    b[operand1-1].set(operand2);
                } else {
                    throw new IllegalArgumentException(String.format("Invalid operation: %s", operation));
                }
    
                System.out.println(b[0].cardinality() + " " + b[1].cardinality());
            }
            
            scanner.close();
        }
    }
    
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    Here is Java BitSet solution - https://programmingoneonone.com/hackerrank-java-bitset-problem-solution.html

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    Java BitSet efficiently manages binary data, similar to how Home Care Services in Honolulu efficiently manage personalized care plans. Both focus on optimizing resources to deliver precise, effective outcomes—whether in programming or providing compassionate support to those in need at home.

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    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.

    Input Format

    The first line contains space-separated integers, (the length of both BitSets and ) and (the number of operations to perform), respectively. The subsequent lines each contain an operation in one of the following forms:

    AND OR XOR FLIP SET In the list above, is the integer or , where denotes and denotes . is an integer denoting a bit's index in the BitSet corresponding to .

    For the binary operations , , and , operands are read from left to right and the BitSet resulting from the operation replaces the contents of the first operand. For example:

    AND 2 1 is the left operand, and is the right operand. This operation should assign the result of to .

    Constraints

    Output Format

    After each operation, print the respective number of set bits in BitSet and BitSet as space-separated integers on a new line.

    Sample Input

    5 4 AND 1 2 SET 1 4 FLIP 2 2 OR 2 1 Sample Output

    0 0 1 0 1 1 1 2 Explanation

    Initially: , , , and . At each step, we print the respective number of set bits in and as a pair of space-separated integers on a new line.

    , The number of set bits in and is .

    Set to (). , . The number of set bits in is and is .

    Flip from () to (). , . The number of set bits in is and is .

    . , . The number of set bits in is and is .

    Language Java 8 More 123456789 } } Line: 9 Col: 2

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    import java.util.BitSet; import java.util.Scanner;

    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 scanner = new Scanner(System.in);
    
        int N = scanner.nextInt();
        int M = scanner.nextInt();
    
        BitSet[] B = new BitSet[]{new BitSet(N), new BitSet(N)};
    
        for( int i=0; i<M; i++ ) {
    
            String cmd = scanner.next();
            int left = scanner.nextInt()-1;
            int right = scanner.nextInt()-1;
    
            switch( cmd ) {
                case "AND" -> B[left].and(B[right]); 
                case "OR" -> B[left].or(B[right]);
                case "XOR" -> B[left].xor(B[right]);
                case "SET" -> B[left].set(right+1); 
                case "FLIP" -> B[left].flip(right+1); 
                default -> {}
            }                        
    
            System.out.print(B[0].cardinality());
            System.out.print(" ");
            System.out.print(B[1].cardinality());
            System.out.println();
        }       
        scanner.close();
    
    }
    

    }