How to turn off a particular bit in a number?

Given a number n and a value k, turn off the kth 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 the expression “~(1 << (k – 1))“, we get a number that has all bits set, except the kth bit. If we do bitwise & of this expression with n, we get a number that has all bits the same as n except the kth bit which is 0. 

Below is the implementation of the 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)

Method 2: Using XOR operator.

Left shift 1 by (k – 1) times and check if kth bit is set or not, if set then take XOR for togging the kth bit.

Below is the implementation of the above approach:

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;

    // Check if the kth bit is set or not
    if (n & (1 << (k - 1))) {

        // Toggle the kth bit
        n = (n ^ (1 << (k - 1)));
    }

    return n;
}

// Driver program to test above function
int main()
{
    int n = 15;
    int k = 4;
    cout << turnOffK(n, k);
    return 0;
}

// This code is contributed by hkdass001

Java

// Java implementation of the approach

import java.util.*;

public class GFG {
    // 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;

        // Check if the kth bit is set or not
        if ((n & (1 << (k - 1))) != 0) {

            // Toggle the kth bit
            n = (n ^ (1 << (k - 1)));
        }

        return n;
    }

    // Driver program to test above function
    public static void main(String[] args)
    {
        int n = 15;
        int k = 4;
        System.out.println(turnOffK(n, k));
    }
}

// This code is contributed by Karandeep1234

Python3

def turn_off_k(n: int, k: int) -> int:
    # k must be greater than 0
    if k <= 0:
        return n

    # Check if the kth bit is set or not
    if n & (1 << (k - 1)):

        # Toggle the kth bit
        n = (n ^ (1 << (k - 1)))

    return n

# Driver program to test above function
if __name__ == '__main__':
    n = 15
    k = 4
    print(turn_off_k(n, k))

C#

// C# implementation of the approach
using System;

public class GFG {

    // 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;

        // Check if the kth bit is set or not
        if ((n & (1 << (k - 1))) != 0) {

            // Toggle the kth bit
            n = (n ^ (1 << (k - 1)));
        }

        return n;
    }
    // Driver Code
    static public void Main()
    {
        int n = 15;
        int k = 4;
        Console.Write(turnOffK(n, k));
        // Code
    }
}
// This code is contributed by Akshay
// Tripathi(akshaytripathi19410)_

Javascript

function turnOffK(n, k) {
    // k must be greater than 0
    if (k <= 0) {
        return n;
    }

    // Check if the kth bit is set or not
    if (n & (1 << (k - 1))) {
        // Toggle the kth bit
        n = (n ^ (1 << (k - 1)));
    }

    return n;
}

// Driver program to test above function
let n = 15;
let k = 4;
console.log(turnOffK(n, k));

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