Bitwise Operators in C - Hacker Rank Solution

CodeWorld19

5 May, 2020

Problem

Objective

This challenge will let you learn about bitwise operators in C.
Inside the CPU, mathematical operations like addition, subtraction, multiplication and division are done in bit-level. To perform bit-level operations in C programming, bitwise operators are used which are explained below.

Bitwise AND operator & The output of bitwise AND is 1 if the corresponding bits of two operands is 1. If either bit of an operand is 0, the result of corresponding bit is evaluated to 0. It is denoted by &. Bitwise OR operator | The output of
bitwise OR is 1 if at least one corresponding bit of two operands is 1. It is denoted by |.
Bitwise XOR (exclusive OR) operator ^ The result of bitwise XOR operator is 1 if the corresponding bits of two operands are opposite. It is denoted by XOR

For example, for integers 3 and 5,

3 = 00000011 (In Binary)
5 = 00000101 (In Binary)

AND operation        OR operation        XOR operation
  00000011             00000011            00000011
& 00000101           | 00000101          ^ 00000101
  ________             ________            ________
  00000001  = 1        00000111  = 7       00000110  = 6

Task

Given set, s= {1,2,3,4......n} find:

the maximum value of a&b which is less than a given integer k , where a and b (where a < b) are two integers from set S.
the maximum value of a|b which is less than a given integer k, where a and b (where a < b) are two integers from set S.
the maximum value of a XOR b which is less than a given integer k, where a and b (where a < b) are two integers from set S.

Input Format

The only line contains 2 space-separated integers n, and k, respectively.

Constraints

2<=n<=10^3
2<=k<=n

Output Format

The first line of output contains the maximum possible value of a & b.
The second line of output contains the maximum possible value of a | b.
The second line of output contains the maximum possible value of a XOR b.

Sample Input :

5 4

Sample Output :

2
3
3

Explanation :

n = 5, k =4
S = {1,2,3,4,5}
All possible values of a and b are:

a = 1, b = 2 ; a&b = 0 a|b = 3 a XOR b = 3
a = 1, b = 3 ; a&b = 1 a|b = 3 a XOR b = 2
a = 1, b = 4 ; a&b = 0 a|b = 5 a XOR b = 5
a = 1, b = 5 ; a&b = 1 a|b = 5 a XOR b = 4
a = 2, b = 3 ; a&b = 2 a|b = 3 a XOR b = 1
a = 2, b = 4 ; a&b = 0 a|b = 6 a XOR b = 6
a = 2, b = 5 ; a&b = 0 a|b = 7 a XOR b = 7
a = 3, b = 4 ; a&b = 0 a|b = 7 a XOR b = 7
a = 3, b = 5 ; a&b = 1 a|b = 7 a XOR b = 6
a = 4, b = 5 ; a&b = 4 a|b = 5 a XOR b = 1

The maximum possible value of a&b that is also <(k=4) is 2, so we print 2 on first line.
The maximum possible value of a|b that is also <(k=4) is 3, so we print 3 on first line.
The maximum possible value of a XOR b that is also <(k=4) is 3, so we print 3 on first line.

Solution :

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main()
{
    int n, k;
    scanf("%d %d", &n, &k);
    int mxAnd = 0, mxOr = 0, mxXor = 0;
    
    for(int i = 1; i <= n; i++){
        for(int j = i + 1; j <= n; j++){
            if(mxAnd < (i & j) && (i & j) < k)
                mxAnd = i & j;
            if(mxOr < (i | j) && (i | j) < k)
                mxOr = i | j;
            if(mxXor < (i ^ j) && (i ^ j) < k)
                mxXor = i ^ j;
        }
    }
    printf("%d\n", mxAnd);
    printf("%d\n", mxOr);
    printf("%d\n", mxXor);
 
    return 0;
}

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