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
Task
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) {
//Write your code here.
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) {
//Write your code here.
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;
}