How to turn off a particular bit in a number?
Last Updated :
28 Feb, 2023
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 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
# 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# 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 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
?>
<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>
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 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
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# 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)_
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));
Time Complexity: O(1)
Auxiliary Space: O(1)
Exercise: Write a function turnOnK() that turns the k'th bit on.
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