How to turn off a particular bit in a number?

Given a number n and a value k, turn off the k’th bit in n. Please note that k = 1 means the rightmost bit.
Examples: 

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

Input:  n = 14, k = 1
Output: 14
The rightmost bit was already off, so no change.

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. 
Below is the implementation of above idea. 

C++

#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;
}

Java

// Java program to turn off a particular bit in a number
import java.io.*;

class TurnOff
{
    // Function to returns a number that has all bits same as n
    // except the k'th bit which is made 0
    static 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
    public static void main (String[] args)
    {
        int n = 15;
        int k = 4;
        System.out.println(turnOffK(n, k));
    }
}
// Contributed by Pramod Kumar

Python3

# Returns a number that
# has all bits same as n
# except the k'th bit
# which is made 0

def turnOffK(n,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 code
n = 15
k = 4
print(turnOffK(n, k))

# This code is contributed
# by Anant Agarwal.

C#

// C# program to turn off a
// particular bit in a number
using System;

class GFG
{

    // Function to returns a number
    // that has all bits same as n
    // except the k'th bit which is
    // made 0
    static 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 Code
    public static void Main ()
    {
        int n = 15;
        int k = 4;
        Console.Write(turnOffK(n, k));
    }
}

// This code is contributed by Nitin Mittal.

PHP

<?php
// PHP program to turn off a
// particular bit in a number

// Returns a number that has
// all bits same as n except
// the k'th bit which is made 0
function turnOffK($n, $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 Code
$n = 15;
$k = 4;
echo turnOffK($n, $k);

// This code is contributed by nitin mittal
?>

Javascript

<script>
// Returns a number that has all bits same as n
// except the k'th bit which is made 0
function turnOffK( n, 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
let n = 15;
let k = 4;
document.write(turnOffK(n, k));

// This code is contributed by rohitsingh07052.
</script>

Output: 

7

Time Complexity: O(1)

Auxiliary Space:  O(1)

Exercise: Write a function turnOnK() that turns the k’th bit on.
This article is contributed by Rahul Jain. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above