Set the rightmost unset bit

Given a non-negative number n. The problem is to set the rightmost unset bit in the binary representation of n. If there are no unset bits, then just leave the number as it is.

Examples:

Input : 21
Output : 23
(21)10 = (10101)2
Rightmost unset bit is at position 2(from right) as 
highlighted in the binary representation of 21.
(23)10 = (10111)2
The bit at position 2 has been set.

Input : 15
Output : 15



Approach: Following are the steps:

  1. If n = 0, return 1.
  2. If all bits of n are set, return n. Refer this post.
  3. Else perform bitwise not on the given number(operation equivalent to 1’s complement). Let it be num = ~n.
  4. Get the position of rightmost set bit of num. Let the position be pos.
  5. Return (1 << (pos – 1)) | n.

C++

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// C++ implementation to set the rightmost unset bit
#include <bits/stdc++.h>
using namespace std;
  
// function to find the position 
// of rightmost set bit
int getPosOfRightmostSetBit(int n)
{
    return log2(n&-n)+1;
}
  
int setRightmostUnsetBit(int n)
{
    // if n = 0, return 1
    if (n == 0)
        return 1;
      
    // if all bits of 'n' are set
    if ((n & (n + 1)) == 0)    
        return n;
      
    // position of rightmost unset bit in 'n'
    // passing ~n as argument
    int pos = getPosOfRightmostSetBit(~n);    
      
    // set the bit at position 'pos'
    return ((1 << (pos - 1)) | n);
}
  
// Driver program to test above
int main()
{
    int n = 21;
    cout << setRightmostUnsetBit(n);
    return 0;

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Java

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// Java implementation to set 
// the rightmost unset bit
  
class GFG {
      
// function to find the position
// of rightmost set bit
static int getPosOfRightmostSetBit(int n) 
{
    return (int)((Math.log10(n & -n)) / (Math.log10(2))) + 1;
}
  
static int setRightmostUnsetBit(int n) 
{
    // if n = 0, return 1
    if (n == 0)
    return 1;
  
    // if all bits of 'n' are set
    if ((n & (n + 1)) == 0)
    return n;
  
    // position of rightmost unset bit in 'n'
    // passing ~n as argument
    int pos = getPosOfRightmostSetBit(~n);
  
    // set the bit at position 'pos'
    return ((1 << (pos - 1)) | n);
}
  
// Driver code
public static void main(String arg[]) {
    int n = 21;
    System.out.print(setRightmostUnsetBit(n));
}
}
  
// This code is contributed by Anant Agarwal.

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Python3

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# Python implementation to
# set the rightmost unset bit
import math
  
# function to find the position 
# of rightmost set bit
def getPosOfRightmostSetBit(n):
  
    return int(math.log2(n&-n)+1)
  
   
def setRightmostUnsetBit(n):
  
    # if n = 0, return 1
    if (n == 0):
        return 1
       
    # if all bits of 'n' are set
    if ((n & (n + 1)) == 0):    
        return n
       
    # position of rightmost unset bit in 'n'
    # passing ~n as argument
    pos = getPosOfRightmostSetBit(~n)    
       
    # set the bit at position 'pos'
    return ((1 << (pos - 1)) | n)
  
# Driver code
  
n = 21
print(setRightmostUnsetBit(n))
  
# This code is contributed
# by Anant Agarwal.

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Output:

23

This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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