Bitwise operator in C Hackerrank 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 s={1,2,3,…,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  which is less than a given integer a^b , 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 < 103
• 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^b.

Sample Input 0

5 4

Sample Output 0

2
3
3

Explanation 0

n = 5,k = 4

s={1,2,3,4,5}

All possible values of a and b are:

1. a=1,b=2;a&b = 0;a|b =3;a^b=3;
2. a=1,b=3;a&b = 1;a|b =3;a^b=2;
3. a=1,b=4;a&b = 0;a|b =5;a^b=5;
4. a=1,b=5;a&b = 1;a|b =5;a^b=4;
5. a=1,b=3;a&b = 2;a|b =3;a^b=1;
6. a=2,b=4;a&b = 0;a|b =6;a^b=6;
7. a=2,b=5;a&b = 0;a|b =7;a^b=7;
8. a=3,b=4;a&b = 0;a|b =7;a^b=7;
9. a=3,b=5;a&b = 1;a|b =7;a^b=6;
10. a=4,b=5;a&b = 4;a|b =5;a^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 second line.
    • The maximum possible value of a^b that is also <(k=4) is 3, so we print  3 on third line.

Code:

#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.
   printf("%d\n",((k-1)|k)<=n? (k-1):(k-2));
    printf("%d\n",((k-1)&(k-2))==0? (k==3?0:(k-2)) : (k-1) );
    printf("%d\n", k-1);
}

int main() {
    int n, k;
  
    scanf("%d %d", &n, &k);
    calculate_the_maximum(n, k);
 
    return 0;
}