Position of rightmost set bit

Write a one-line function to return the position of the first 1 from right to left, in the binary representation of an Integer. 

I/P    18,   Binary Representation 010010
O/P   2
I/P    19,   Binary Representation 010011
O/P   1

Algorithm: (Example 12(1100))
Let I/P be 12 (1100)
1. Take two’s complement of the given no as all bits are reverted
except the first ‘1’ from right to left (0100)
2  Do a bit-wise & with original no, this will return no with the
required one only (0100)
3  Take the log2 of the no, you will get (position – 1) (2)
4  Add 1 (3)

Explanation –

(n&~(n-1)) always return the binary number containing the rightmost set bit as 1.
if N = 12 (1100) then it will return 4 (100)
Here log2 will return you, the number of times we can express that number in a power of two.
For all binary numbers containing only the rightmost set bit as 1 like 2, 4, 8, 16, 32….
We will find that position of rightmost set bit is always equal to log2(Number)+1

Program:

C++

// C++ program for Position
// of rightmost set bit
#include <iostream>
#include <math.h>
using namespace std;

class gfg
{

public:
unsigned int getFirstSetBitPos(int n)
{
    return log2(n & -n) + 1;
}
};

// Driver code
int main()
{
    gfg g;
    int n = 12;
    cout << g.getFirstSetBitPos(n);
    return 0;
}

// This code is contributed by SoM15242

C

// C program for Position
// of rightmost set bit
#include <math.h>
#include <stdio.h>

unsigned int getFirstSetBitPos(int n)
{
    return log2(n & -n) + 1;
}

// Driver code
int main()
{
    int n = 12;
    printf("%u", getFirstSetBitPos(n));
    getchar();
    return 0;
}

Java

// Java Code for Position of rightmost set bit
class GFG {

    public static int getFirstSetBitPos(int n)
    {
        return (int)((Math.log10(n & -n)) / Math.log10(2)) + 1;
    }

    // Drive code
    public static void main(String[] args)
    {
        int n = 12;
        System.out.println(getFirstSetBitPos(n));
    }
}
// This code is contributed by Arnav Kr. Mandal

Python3

# Python Code for Position
# of rightmost set bit

import math

def getFirstSetBitPos(n):

     return math.log2(n&-n)+1

# driver code

n = 12
print(int(getFirstSetBitPos(n)))

# This code is contributed
# by Anant Agarwal.

C#

// C# Code for Position of rightmost set bit
using System;

class GFG {
    public static int getFirstSetBitPos(int n)
    {
        return (int)((Math.Log10(n & -n))
                / Math.Log10(2)) + 1;
    }

    // Driver method
    public static void Main()
    {
        int n = 12;
        Console.WriteLine(getFirstSetBitPos(n));
    }
}

// This code is contributed by Sam007

PHP

<?php
// PHP Code for Position of
// rightmost set bit

function getFirstSetBitPos($n)
{
    return ceil(log(($n& -
                     $n) + 1, 2));
}

// Driver Code
$n = 12;
echo getFirstSetBitPos($n);

// This code is contributed by m_kit
?>

Javascript

<script>
// JavaScript program for Position
// of rightmost set bit

function getFirstSetBitPos(n)
{
    return Math.log2(n & -n) + 1;
}

// Driver code
    let g;
    let n = 12;
    document.write(getFirstSetBitPos(n));

// This code is contributed by Manoj.
</script>

Output

3

Time Complexity: O(log₂n)
Auxiliary Space: O(1)

Using ffs() function: ffs() function returns the index of first least significant set bit. The indexing starts in ffs() function from 1. 
For example: 
n = 12 = 1100 

In the above example, ffs(n) returns the rightmost set bit index which is 3. 

C++

// C++ program to find the
// position of first rightmost
// set bit in a given number.
#include <bits/stdc++.h>
using namespace std;

// Function to find rightmost
// set bit in given number.
int getFirstSetBitPos(int n)
{
    return ffs(n);
}

// Driver function
int main()
{
    int n = 12;
    cout << getFirstSetBitPos(n) << endl;
    return 0;
}

Java

// Java program  to find the
// position of first rightmost
// set bit in a given number
import java.util.*;

class GFG{

// Function to find rightmost
// set bit in given number.
static int getFirstSetBitPos(int n)
{
    return (int)((Math.log10(n & -n)) / Math.log10(2)) + 1;
}

// Driver code
public static void main(String[] args)
{
    int n = 12;
    System.out.print( getFirstSetBitPos(n));

}
}

// This code is contributed by sanjoy_62.

Python3

# Python3 program to find the
# position of first rightmost
# set bit in a given number
import math

# Function to find rightmost
# set bit in given number.
def getFirstSetBitPos(n):

    return int(math.log2 (n & -n) + 1)

# Driver Code
if __name__ == '__main__':

    n = 12
    print(getFirstSetBitPos(n))

# This code is contributed by nirajgusain5.

C#

// C# program  to find the
// position of first rightmost
// set bit in a given number
using System;
public class GFG{

// Function to find rightmost
// set bit in given number.
static int getFirstSetBitPos(int n)
{
    return (int)((Math.Log10(n & -n)) / Math.Log10(2)) + 1;
}

// Driver code
public static void Main(String[] args)
{
    int n = 12;
    Console.Write( getFirstSetBitPos(n));

}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// Javascript program to find the
// position of first rightmost
// set bit in a given number

// Function to find rightmost
// set bit in given number.
function getFirstSetBitPos(n)
{
    return Math.log2(n & -n) + 1;
}

// Driver Code

    let n = 12;
    document.write( getFirstSetBitPos(n));

</script>

Output

3

Time Complexity: O(log₂n)
Auxiliary Space: O(1)

Using XOR and & operator : 
Initialize m as 1 as check its XOR with the bits starting from the rightmost bit. Left shift m by one till we find the first set bit, as the first set bit gives a number when we perform a & operation with m. 

C++

// C++ program to find the first
// rightmost set bit using XOR operator
#include <bits/stdc++.h>
using namespace std;

// function to find the rightmost set bit
int PositionRightmostSetbit(int n)
{
      if(n == 0)
          return 0;
    // Position variable initialize with 1
    // m variable is used to check the set bit
    int position = 1;
    int m = 1;

    while (!(n & m)) {

        // left shift
        m = m << 1;
        position++;
    }
    return position;
}
// Driver Code
int main()
{
    int n = 16;
    // function call
    cout << PositionRightmostSetbit(n);
    return 0;
}

Java

// Java program to find the
// first rightmost set bit
// using XOR operator

class GFG {

    // function to find
    // the rightmost set bit
    static int PositionRightmostSetbit(int n)
    {
        // Position variable initialize
        // with 1 m variable is used to
        // check the set bit
        int position = 1;
        int m = 1;

        while ((n & m) == 0) {

            // left shift
            m = m << 1;
            position++;
        }
        return position;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int n = 16;

        // function call
        System.out.println(PositionRightmostSetbit(n));
    }
}

// This code is contributed
// by Smitha

Python3

# Python3 program to find
# the first rightmost set
# bit using XOR operator

# function to find the
# rightmost set bit
def PositionRightmostSetbit(n):

    # Position variable initialize
    # with 1 m variable is used
    # to check the set bit
    position = 1
    m = 1

    while (not(n & m)) :

        # left shift
        m = m << 1
        position += 1

    return position

# Driver Code
n = 16

# function call
print(PositionRightmostSetbit(n))

# This code is contributed
# by Smitha

C#

// C# program to find the
// first rightmost set bit
// using XOR operator
using System;

class GFG {

    // function to find
    // the rightmost set bit
    static int PositionRightmostSetbit(int n)
    {
        // Position variable initialize
        // with 1 m variable is used to
        // check the set bit
        int position = 1;
        int m = 1;

        while ((n & m) == 0) {

            // left shift
            m = m << 1;
            position++;
        }
        return position;
    }

    // Driver Code
    static public void Main()
    {
        int n = 16;

        // function call
        Console.WriteLine(
            PositionRightmostSetbit(n));
    }
}

// This code is contributed
// by @ajit

PHP

<?php
// PHP program to find the
// first rightmost set bit
// using XOR operator

// function to find the
// rightmost set bit
function PositionRightmostSetbit($n)
{
    // Position variable initialize
    // with 1 m variable is used to
    // check the set bit
    $position = 1;
    $m = 1;

    while (!($n & $m))
    {

        // left shift
        $m = $m << 1;
        $position++;
    }
    return $position;
}

// Driver Code
$n = 16;

// function call
echo PositionRightmostSetbit($n);

// This code is contributed by ajit
?>

Javascript

<script>

    // Javascript program to find the first
    // rightmost set bit using XOR operator

    // function to find the rightmost set bit
    function PositionRightmostSetbit(n)
    {

        // Position variable initialize with 1
        // m variable is used to check the set bit
        let position = 1;
        let m = 1;

        while ((n & m) == 0) {

            // left shift
            m = m << 1;
            position++;
        }
        return position;
    }

    let n = 16;

    // function call
    document.write(PositionRightmostSetbit(n));

    // This code is contributed by divyesh072019.
</script>

Output

5

Time Complexity: O(log₂n)
Auxiliary Space: O(1)

This approach has been contributed by mubashshir ahmad

Using Left Shift (<<) : Initialize pos with 1, iterate up to INT_SIZE(Here 32) and check whether bit is set or not, if bit is set then break the loop, else increment the pos.  

C++

// C++ implementation of above approach
#include <iostream>
using namespace std;
#define INT_SIZE 32

int Right_most_setbit(int num)
{
  if(num==0)// for num==0 there is zero set bit
  {  return 0;
  }
  else
  {
    int pos = 1;
    // counting the position of first set bit
    for (int i = 0; i < INT_SIZE; i++) {
        if (!(num & (1 << i)))
            pos++;
        else
            break;
    }
    return pos;
}
}
int main()
{
    int num = 0;
    int pos = Right_most_setbit(num);
    cout << pos << endl;
    return 0;
}
// This approach has been contributed by @yashasvi7

Java

//Java implementation of above approach
public class GFG {

    static int INT_SIZE = 32;

    static int Right_most_setbit(int num)
    {
        int pos = 1;
        // counting the position of first set bit
        for (int i = 0; i < INT_SIZE; i++) {
            if ((num & (1 << i))== 0)
                pos++;

            else
                break;
        }
        return pos;
    }

    //Driver code
    public static void main(String[] args) {

         int num = 18;
            int pos = Right_most_setbit(num);
            System.out.println(pos);
    }
}

Python3

# Python 3 implementation of above approach

INT_SIZE = 32

def Right_most_setbit(num) :

    pos = 1

    # counting the position of first set bit
    for i in range(INT_SIZE) :
        if not(num & (1 << i)) :
            pos += 1
        else :
            break

    return pos



if __name__ == "__main__" :

    num = 18
    pos = Right_most_setbit(num)
    print(pos)

# This code is contributed by ANKITRAI1

C#

// C# implementation of above approach
using System;

class GFG {

    static int INT_SIZE = 32;

    static int Right_most_setbit(int num)
    {
        int pos = 1;

        // counting the position
        // of first set bit
        for (int i = 0; i < INT_SIZE; i++)
        {
            if ((num & (1 << i))== 0)
                pos++;

            else
                break;
        }
        return pos;
    }

    // Driver code
    static public void Main ()
    {
        int num = 18;
        int pos = Right_most_setbit(num);
        Console.WriteLine(pos);
    }
}

PHP

<?php
// PHP implementation of above approach

function Right_most_setbit($num)
{
    $pos = 1;
    $INT_SIZE = 32;

    // counting the position
    // of first set bit
    for ($i = 0; $i < $INT_SIZE; $i++)
    {
        if (!($num & (1 << $i)))
            $pos++;
        else
            break;
    }
    return $pos;
}

// Driver code
$num = 18;
$pos = Right_most_setbit($num);
echo $pos;
echo ("\n")

// This code is contributed
// by Shivi_Aggarwal
?>

Javascript

<script>

// Javascript implementation of above approach
let INT_SIZE = 32;

function Right_most_setbit(num)
{
    let pos = 1;

    // Counting the position of first set bit
    for(let i = 0; i < INT_SIZE; i++)
    {
        if ((num & (1 << i)) == 0)
            pos++;
        else
            break;
    }
    return pos;
}

// Driver code
let num = 18;
let pos = Right_most_setbit(num);

document.write(pos);

// This code is contributed by mukesh07

</script>

Output : 

 2

Time Complexity: O(1)
Auxiliary Space: O(1)

Another Method Using right Shift(>>):

Initialize pos=1. Iterate till number>0,  at each step check if the last bit is set. If last bit is set, return current position, else increment pos by 1 and right shift n by 1.

C++

// C++ program for above approach
#include<bits/stdc++.h>
using namespace std;

// Program to find position of
// rightmost set bit
int PositionRightmostSetbit(int n)
{
  int p=1;

  // Iterate till number>0
  while(n > 0)
  {

    // Checking if last bit is set
    if(n&1){
      return p;
    }

    // Increment position and right shift number
    p++;
    n=n>>1;
  }

  // set bit not found.
  return -1;
}

// Driver Code
int main()
{
  int n=18;

  // Function call
  int pos=Last_set_bit(n);

  if(pos!=-1)
    cout<<pos;
  else
    cout<<0;

  return 0;
}

Java

// Java program for above approach
import java.io.*;

class GFG{

// Function to find position of
// rightmost set bit
public static int Last_set_bit(int n)
{
    int p = 1;

    // Iterate till number>0
    while (n > 0)
    {

        // Checking if last bit is set
        if ((n & 1) > 0)
        {
            return p;
        }

        // Increment position and
        // right shift number
        p++;
        n = n >> 1;
    }

    // set bit not found.
    return -1;
}

// Driver Code
public static void main(String[] args)
{
    int n = 18;

    // Function call
    int pos = Last_set_bit(n);

    if (pos != -1)
        System.out.println(pos);
    else
        System.out.println("0");
}
}

// This code is contributed by RohitOberoi

Python3

# Python program for above approach

# Program to find position of
# rightmost set bit
def PositionRightmostSetbit( n):
  p = 1

  # Iterate till number>0
  while(n > 0):

    # Checking if last bit is set
    if(n&1):
      return p

    # Increment position and right shift number
    p += 1
    n = n>>1

  # set bit not found.
  return -1;

# Driver Code
n = 18
# Function call
pos = PositionRightmostSetbit(n)
if(pos != -1):
  print(pos)
else:
  print(0)

# This code is contributed by rohitsingh07052.

C#

// C# program for above approach
using System;

class GFG{

// Function to find position of
// rightmost set bit
public static int Last_set_bit(int n)
{
    int p = 1;

    // Iterate till number>0
    while (n > 0)
    {

        // Checking if last bit is set
        if ((n & 1) > 0)
        {
            return p;
        }

        // Increment position and
        // right shift number
        p++;
        n = n >> 1;
    }

    // Set bit not found.
    return -1;
}

// Driver Code
public static void Main(string[] args)
{
    int n = 18;

    // Function call
    int pos = Last_set_bit(n);

    if (pos != -1)
        Console.WriteLine(pos);
    else
        Console.WriteLine("0");
}
}

// This code is contributed by AnkThon

Javascript

<script>

// Javascript program for above approach

// Program to find position of
// rightmost set bit
function Last_set_bit(n)
{
    let p = 1;

    // Iterate till number>0
    while (n > 0)
    {

        // Checking if last bit is set
        if ((n & 1) > 0)
        {
            return p;
        }

        // Increment position and
        // right shift number
        p++;
        n = n >> 1;
    }

    // Set bit not found.
    return -1;
}

// Driver code
let n = 18;

// Function call
let pos = Last_set_bit(n);

if (pos != -1)
{
    document.write(pos);
}
else
{
    document.write(0);
}

// This code is contributed by divyeshrabadiya07

</script>

Output

2

Time Complexity: O(log₂n)
Auxiliary Space: O(1)