Bitwise Operators in C – Hacker Rank Solution ~ Multiplex Coder

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.

Bitwise Operators in C – Hacker Rank Solution

Code:

#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;

}

Disclaimer: The above Problem (Bitwise Operators in C) is generated by Hackerrank but the Solution is Provided by MultiplexCoder. This tutorial is only for Educational and Learning purposes. Authority if any of the queries regarding this post or website fill the following contact form thank you.

Bitwise Operators in C – Hacker Rank Solution

Objective

Task

Input Format

Constraints

Output Format

Sample Input :

Sample Output :

Explanation :

Bitwise Operators in C – Hacker Rank Solution

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