Hackerrank Bitwise Operators Solution

# Hackerrank Bitwise Operators Solution

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 .

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

Given set , find:

• the maximum value of  which is less than a given integer , where  and  (where ) are two integers from set .
• the maximum value of  which is less than a given integer , where  and  (where ) are two integers from set .
• the maximum value of  which is less than a given integer , where  and  (where ) are two integers from set .

Input Format.

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

Constraints.

Output Format.

• The first line of output contains the maximum possible value of .
• The second line of output contains the maximum possible value of .
• The second line of output contains the maximum possible value of .

Sample Input 0

5 4

Sample Output 0.

233

Explanation 0.

All possible values of  and  are:

• The maximum possible value of  that is also  is , so we print  on first line.
• The maximum possible value of  that is also  is , so we print  on second line.
• The maximum possible value of  that is also  is , so we print  on third line.

### Solution in c

Approach 1.

#include<stdio.h>
int main()
{
int i,j,n,k,p=0,q=0,r=0;
scanf("%d %d",&n,&k);
for(i=1;i<n;i++)
{
for(j=i+1;j<=n;j++)
{
if((i&j) < k && p < (i&j))
{
p=i&j;
}
if((i|j) < k && q < (i|j))
{
q=i|j;
}
if((i^j) < k && r < (i^j))
{
r=i^j;
}
}

}
printf("%d\n%d\n%d",p,q,r);

}

Approach 2.

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
//Complete the following function.

void calculate_the_maximum(int n, int k) {
int and, or, xor, i, j, temp;
and = 0;
or = 0;
xor = 0;
for(i=1; i<=n; i++ ){
for(j=i+1; j<=n; j++){
temp = i&j;
if(temp>and && temp<k){and = temp;}
temp = i|j;
if(temp>or && temp<k){or = temp;}
temp =i^j;
if(temp>xor && temp<k){xor = temp;}
}
}
printf("%d\n%d\n%d", and, or, xor);
}

int main() {
int n, k;

scanf("%d %d", &n, &k);
calculate_the_maximum(n, k);

return 0;
}


Approach 3.

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
//Complete the following function.

void calculate_the_maximum(int n, int k) {
int i=0,j=0;
int andmax=0,ormax=0,xormax=0;
for(i=1;i<n;i++)
{
for(j=i+1;j<=n;j++)
{
if((i & j)>andmax && (i & j)<k)
andmax=(i & j);
if((i | j)>ormax && (i | j)<k)
ormax=(i | j);
if((i ^ j)>xormax && (i ^ j)<k)
xormax=(i ^ j);
}
}
printf("%d\n%d\n%d",andmax,ormax,xormax);
}

int main() {
int n, k;

scanf("%d %d", &n, &k);
calculate_the_maximum(n, k);

return 0;
}