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Set the rightmost unset bit
  • Difficulty Level : Easy
  • Last Updated : 17 May, 2021

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 to 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++




// 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;

Java




// 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.

Python3




# Python3 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.

C#




// C# implementation to set 
// the rightmost unset bit
using System;
  
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;
      
    Console.Write(setRightmostUnsetBit(n));
}
}
  
// This code is contributed by shivanisinghss2110

Javascript




<script>
  
// JavaScript implementation to set
// the rightmost unset bit
      
    // function to find the position
    // of rightmost set bit
    function getPosOfRightmostSetBit(n)
    {
        return ((Math.log10(n & -n)) / (Math.log10(2))) + 1;
    }
      
    function 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
    let pos = getPosOfRightmostSetBit(~n);
   
    // set the bit at position 'pos'
    return ((1 << (pos - 1)) | n);
    }
      
    // Driver code
    let n = 21;
    document.write(setRightmostUnsetBit(n));
      
  
// This code is contributed by unknown2108
  
</script>

Output: 

23

Alternate Implementation
The idea is to use Integer.toBinaryString() 



Java




import java.io.*;
import java.util.Scanner;
  
public class SetMostRightUnsetBit {
    public static void main(String args[])
    {
        int a = 21;
        setMostRightUnset(a);
    }
  
    private static void setMostRightUnset(int a)
    {
        // will get a number with all set bits except the 
        // first set bit
        int x = a ^ (a - 1);
        System.out.println(Integer.toBinaryString(x));
  
        // We reduce it to the number with single 1's on 
        // the position of first set bit in given number
        x = x & a; 
        System.out.println(Integer.toBinaryString(x));
  
        // Move x on right by one shift to make OR 
        // operation and make first rightest unset bit 1
        x = x >> 1
  
        int b = a | x;
  
        System.out.println("before setiing bit "
                               Integer.toBinaryString(a));
        System.out.println("after setting bit "
                               Integer.toBinaryString(b));
    }
}

Python3




def setMostRightUnset(a):
      
    # Will get a number with all set 
    # bits except the first set bit
    x = a ^ (a - 1)
    print(bin(x)[2:]) 
  
    # We reduce it to the number with 
    # single 1's on the position of 
    # first set bit in given number
    x = x & a
    print(bin(x)[2:]) 
  
    # Move x on right by one shift to 
    # make OR operation and make first 
    # rightest unset bit 1
    x = x >> 1
  
    b = a | x
  
    print("before setiing bit ", bin(a)[2:]) 
    print("after setting bit ", bin(b)[2:]) 
      
# Driver Code
if __name__ == '__main__':
      
    a = 21
      
    setMostRightUnset(a)
  
# This code is contributed by mohit kumar 29

C#




using System;
  
class SetMostRightUnsetBit{
      
// Driver Code
public static void Main(String []args)
{
    int a = 21;
    setMostRightUnset(a);
}
  
private static void setMostRightUnset(int a)
{
      
    // will get a number with all set bits 
    // except the first set bit
    int x = a ^ (a - 1);
    Console.WriteLine(Convert.ToString(x));
  
    // We reduce it to the number with single 1's on 
    // the position of first set bit in given number
    x = x & a; 
    Console.WriteLine(Convert.ToString(x));
  
    // Move x on right by one shift to make OR 
    // operation and make first rightest unset bit 1
    x = x >> 1; 
  
    int b = a | x;
  
    Console.WriteLine("before setiing bit " +
    Convert.ToString(a, 2));
    Console.WriteLine("after setting bit " +
    Convert.ToString(b, 2));
}
}
  
// This code is contributed by shivanisinghss2110

Javascript




<script>
      
    function setMostRightUnset(a)
    {
        // will get a number with all set bits except the
        // first set bit
        let x = a ^ (a - 1);
        document.write((x >>> 0).toString(2)+"<br>");
   
        // We reduce it to the number with single 1's on
        // the position of first set bit in given number
        x = x & a;
        document.write((x >>> 0).toString(2)+"<br>");
   
        // Move x on right by one shift to make OR
        // operation and make first rightest unset bit 1
        x = x >> 1;
   
        let b = a | x;
   
        document.write("before setiing bit " +
                               (a >>> 0).toString(2)+"<br>");
        document.write("after setting bit " +
                               (b >>> 0).toString(2)+"<br>");
    }
      
    let a = 21;
    setMostRightUnset(a);
  
  
// This code is contributed by patel2127
</script>

Output:

1
1
before setiing bit 10101
after setting bit 10101

This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.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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