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



Alternate Implementation in Java

The idea is to use Integer.toBinaryString()











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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)); 


    } 


} 



















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