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 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()

C++

// C++ program to implement the approach
#include <bits/stdc++.h>
using namespace std;

// function to binary string and remove
// the leading zeroes
string getBinary(int n)
{
  string binary = bitset<32>(n).to_string();
  binary.erase(0, binary.find_first_not_of('0'));
  return binary;
}

void setMostRightUnset(int a)
{

  // will get a number with all set bits except the
  // first set bit
  int x = a ^ (a - 1);
  cout << getBinary(x) << endl;

  // We reduce it to the number with single 1's on
  // the position of first set bit in given number
  x = x & a;
  cout << getBinary(x) << endl;

  // 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;

  cout << "before setting bit " << getBinary(a) << endl;
  cout << "after setting bit " << getBinary(b) << endl;
}

int main()
{
  int a = 21;
  setMostRightUnset(a);

  return 0;
}

// This code is contributed by phasing17

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 setting 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 setting 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 setting 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 setting 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 setting bit 10101
after setting bit 10101

Another Approach:

The idea is x|(x+1) sets the lowest(rightmost) unset bit but it may set a leading zero bit if the given number doesn’t contain any unset bits , for example 

if all bits set
15 is  1111 = 00001111 
16 is 10000 = 00010000
             _________
x|(x+1)          00011111
on performing x|(x+1) it converts to 00011111 
but here we should only consider only 1111 and return 1111 as there are no unset bits.

else 
13 is 1101 = 00001101
14 is 1110 = 00001110
            _________
x|(x+1)         00001111
on performing x|(x+1) it converts to 00011111 which is required number.

So we can eliminate this type of redundant operation by checking if there are any zeroes in the given number. We observe only number whose all bits set are of form 2^k -1 have no unset bits in them so by catching these number we can perform our idea.

C++

// C++ implementation to set the rightmost unset bit
#include <bits/stdc++.h>
using namespace std;

int setRightmostUnsetBit(int n)
{
    // if all bits of 'n' are set
    // the number is of form 2^k -1 return n
    if (!(n & (n + 1)))
        return n;
    // else
    return n | (n + 1);
}

// Driver program to test above
int main()
{
    int n = 21;
    cout << setRightmostUnsetBit(n);
    return 0;
}

// This code is contributed by Kasina Dheeraj.

Java

// Java implementation to set
// the rightmost unset bit

class GFG {

    static int setRightmostUnsetBit(int n)
    {
        // if all bits of 'n' are set
        // the number is of form 2^k -1 return n
        if ((n & (n + 1)) == 0)
            return n;

        // else
        return n | (n + 1);
    }

    // Driver code
    public static void main(String arg[])
    {
        int n = 21;
        System.out.print(setRightmostUnsetBit(n));
    }
}

// This code is contributed by Kasina Dheeraj.

Python3

# Python3 implementation to
# set the rightmost unset bit
import math


def setRightmostUnsetBit(n):

    # if all bits of 'n' are set
    # the number is of form 2^k -1 return n
    if ((n & (n + 1)) == 0):
        return n
    # else
    return n | (n+1)

# Driver code


n = 21
print(setRightmostUnsetBit(n))

# This code is contributed by Kasina Dheeraj.

C#

// C# implementation to set
// the rightmost unset bit
using System;

class GFG {

    static int setRightmostUnsetBit(int n)
    {
        // if all bits of 'n' are set
        // the number is of form 2^k -1 return n
        if ((n & (n + 1)) == 0)
            return n;
        // else
        return n | (n + 1);
    }

    // Driver code
    public static void Main(String[] arg)
    {
        int n = 21;

        Console.Write(setRightmostUnsetBit(n));
    }
}

// This code is contributed by Kasina Dheeraj.

Javascript

// JavaScript implementation to set the rightmost unset bit

function setRightmostUnsetBit(n)
{
    // if all bits of 'n' are set
    // the number is of form 2^k -1 return n
    if (!(n & (n + 1)))
        return n;
    // else
    return n | (n + 1);
}

// Driver program to test above
let n = 21;
console.log(setRightmostUnsetBit(n));

// This code is contributed by phasing17

Output:

23

Time complexity: O(1)

Space Complexity: O(1)

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