Difficulty Level: Rookie

Given a number n and a value k, turn of the k’th bit in n.

Examples:

Input:  n = 15, k = 1
Output: 14

Input:  n = 15, k = 2
Output: 13

Input:  n = 15, k = 3
Output: 11

Input:  n = 15, k = 4
Output: 7

Input:  n = 15, k >= 5
Output: 15

The idea is to use bitwise <<, & and ~ operators. Using expression “~(1 << (k – 1))“, we get a number which has all bits set, except the k’th bit. If we do bitwise & of this expression with n, we get a number which has all bits same as n except the k’th bit which is 0.

Following is C++ implementation of this.

#include <iostream>
using namespace std;
 
// Returns a number that has all bits same as n
// except the k'th bit which is made 0
int turnOffK(int n, int k)
{
    // k must be greater than 0
    if (k <= 0) return n;
 
    // Do & of n with a number with all set bits except
    // the k'th bit
    return (n & ~(1 << (k - 1)));
}
 
// Driver program to test above function
int main()
{
    int n = 15;
    int k = 4;
    cout << turnOffK(n, k);
    return 0;
}

Output:

7