Skip to content
Related Articles

Related Articles

Position of rightmost different bit

View Discussion
Improve Article
Save Article
  • Difficulty Level : Easy
  • Last Updated : 12 Sep, 2022
View Discussion
Improve Article
Save Article

Given two numbers m and n. Find the position of the rightmost different bit in the binary representation of numbers. It is guaranteed that such a bit exists

Examples: 

Input: m = 11, n = 9
Output: 2
Explanation: 
(11)10 = (1011)2
(9)10 = (1001)2
It can be seen that 2nd bit from the right is different 

Input: m = 52, n = 4
Output: 5
Explanation: 
(52)10 = (110100)2
(4)10 = (100)2, can also be written as = (000100)2
It can be seen that 5th bit from the right is different

Recommended Practice

Position of rightmost different bit using XOR:

Get the bitwise xor of m and n. Let it be xor_value = m ^ n. Now, find the position of rightmost set bit in xor_value. As 0 XOR 1 and 1 XOR 0 equals 1, so if a bit is set in the XOR value then it means that the bits at that position were different in the given numbers

An efficient way to find the rightmost set bit: 

log2(n & -n) + 1 gives us the position of the rightmost set bit.
(-n) reverses all the bits from left to right till the last set bit

for example: n = 16810
binary signed 2’s complement of n  = 00000000101010002
binary signed 2’s complement of -n = 11111111010110002
∴ (n & -n) = 00000000000010002 = 8
now, log2(n & -n) = log2(8) = 3
log2(n & -n) + 1 = 4 (position of rightmost set bit)

Below is the implementation of the above approach:

C++




// C++ implementation to find the position
// of rightmost different bit
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the position of
// rightmost set bit in 'n'
// returns 0 if there is no set bit.
int getRightMostSetBit(int n)
{
    // to handle edge case when n = 0.
    if (n == 0)
        return 0;
 
    return log2(n & -n) + 1;
}
 
// Function to find the position of
// rightmost different bit in the
// binary representations of 'm' and 'n'
// returns 0 if there is no
// rightmost different bit.
int posOfRightMostDiffBit(int m, int n)
{
    // position of rightmost different
    //  bit
 
    return getRightMostSetBit(m ^ n);
}
 
// Driver code
int main()
{
    int m = 52, n = 24;
 
    // Function call
    cout << "Position of rightmost different bit:"
         << posOfRightMostDiffBit(m, n) << endl;
    return 0;
}


Java




// Java implementation to find the position
// of rightmost different bit
 
class GFG {
 
    // Function to find the position of
    // rightmost set bit in 'n'
    // return 0 if there is no set bit.
    static int getRightMostSetBit(int n)
    {
        if (n == 0)
            return 0;
 
        return (int)((Math.log10(n & -n)) / Math.log10(2))
            + 1;
    }
 
    // Function to find the position of
    // rightmost different bit in the
    // binary representations of 'm' and 'n'
    static int posOfRightMostDiffBit(int m, int n)
    {
        // position of rightmost different bit
        return getRightMostSetBit(m ^ n);
    }
 
    // Driver code
    public static void main(String arg[])
    {
        int m = 52, n = 4;
 
        // Function call
        System.out.print("Position = "
                         + posOfRightMostDiffBit(m, n));
    }
}
 
// This code is contributed by Anant Agarwal.


Python3




# Python implementation
# to find the position
# of rightmost different bit
 
import math
 
# Function to find the position of
# rightmost set bit in 'n'
 
 
def getRightMostSetBit(n):
    if (n == 0):
        return 0
 
    return math.log2(n & -n) + 1
 
 
# Function to find the position of
# rightmost different bit in the
# binary representations of 'm' and 'n'
def posOfRightMostDiffBit(m, n):
 
    # position of rightmost different
    # bit
    return getRightMostSetBit(m ^ n)
 
 
# Driver code
if __name__ == "__main__":
    m = 52
    n = 4
 
    # Function call
    print("position = ", int(posOfRightMostDiffBit(m, n)))
 
# This code is contributed
# by Anant Agarwal.


C#




// C# implementation to find the position
// of rightmost different bit
using System;
 
class GFG {
 
    // Function to find the position of
    // rightmost set bit in 'n'
    static int getRightMostSetBit(int n)
    {
        if (n == 0)
            return 0;
        return (int)((Math.Log10(n & -n)) / Math.Log10(2))
            + 1;
    }
 
    // Function to find the position of
    // rightmost different bit in the
    // binary representations of 'm' and 'n'
    static int posOfRightMostDiffBit(int m, int n)
    {
        // position of rightmost different bit
        return getRightMostSetBit(m ^ n);
    }
 
    // Driver code
    public static void Main()
    {
        int m = 52, n = 4;
 
        // Function call
        Console.Write("Position = "
                      + posOfRightMostDiffBit(m, n));
    }
}
 
// This code is contributed by Smitha.


PHP




<?php
// PHP implementation to
// find the position of
// rightmost different bit
 
// Function to find the position
// of rightmost set bit in 'n'
function getRightMostSetBit($n)
{
    if ($n == 0)
      return 0;
    return log($n & -$n, (2)) + 1;
}
 
// Function to find the position of
// rightmost different bit in the
// binary representations of 'm'
// and 'n'
function posOfRightMostDiffBit($m, $n)
{
     
    // position of rightmost
    // different bit
    return getRightMostSetBit($m ^ $n);
}
 
    // Driver Code
    $m = 52;
    $n = 4;
 
    // Function call
    echo posOfRightMostDiffBit($m, $n);
         
// This code is contributed by Ajit
?>


Javascript




<script>
// JavaScript implementation to find the position
// of rightmost different bit
 
// Function to find the position of
// rightmost set bit in 'n'
// returns 0 if there is no set bit.
function getRightMostSetBit(n)
{
 
    // to handle edge case when n = 0.
    if (n == 0)
        return 0;
     
    return Math.log2(n & -n) + 1;
}
 
// Function to find the position of
// rightmost different bit in the
// binary representations of 'm' and 'n'
// returns 0 if there is no
// rightmost different bit.
function posOfRightMostDiffBit(m, n)
{
    // position of rightmost different
    // bit
     
    return getRightMostSetBit(m ^ n);
}
 
// Driver program
 
    let m = 52, n = 24;
 
    document.write("Position of rightmost different bit:"
        + posOfRightMostDiffBit(m, n) + "<br>");
 
 
 
// This code is contributed by Surbhi Tyagi.
</script>


Output

Position of rightmost different bit:3

Time Complexity: O(log2 N), this time complexity is equal to O(1) as one has to check for at most 31 bits only
Auxiliary Space: O(1)

Position of rightmost different bit using ffs() function:

ffs() function searches the first set bit from the right side and then returns the index of that bit (1-based indexing). So we can use this function on the XOR of both the values to get the index of the rightmost different bit

Below is the implementation of the above approach:

C++




// C++ implementation to find the
// position of rightmost different
// bit in two number.
#include <bits/stdc++.h>
using namespace std;
 
// function to find rightmost different
//  bit in two numbers.
int posOfRightMostDiffBit(int m, int n)
{
    return ffs(m ^ n);
}
 
// Driver code
int main()
{
    int m = 52, n = 4;
 
    // Function call
    cout << "Position = " << posOfRightMostDiffBit(m, n);
    return 0;
}


Java




// Java implementation to find the
// position of rightmost different
// bit in two number.
import java.util.*;
class GFG {
 
    // function to find rightmost
    // different bit in two numbers.
    static int posOfRightMostDiffBit(int m, int n)
    {
        return (int)Math.floor(
                   Math.log10(Math.pow(m ^ n, 2)))
            + 2;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int m = 52, n = 4;
 
        // Function call
        System.out.println("Position = "
                           + posOfRightMostDiffBit(m, n));
    }
}
 
// This code is contributed by gauravrajput1


Python3




# Python3 implementation to find the
# position of rightmost different
# bit in two number.
from math import floor, log10
 
# Function to find rightmost different
# bit in two numbers.
 
 
def posOfRightMostDiffBit(m, n):
 
    return floor(log10(pow(m ^ n, 2))) + 2
 
 
# Driver code
if __name__ == '__main__':
 
    m, n = 52, 4
 
    # Function call
    print("Position = ",
          posOfRightMostDiffBit(m, n))
 
# This code is contributed by mohit kumar 29


C#




// C# implementation to find the
// position of rightmost different
// bit in two number.
using System;
class GFG {
 
    // function to find rightmost
    // different bit in two numbers.
    static int posOfRightMostDiffBit(int m, int n)
    {
        return (int)Math.Floor(
                   Math.Log10(Math.Pow(m ^ n, 2)))
            + 2;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int m = 52, n = 4;
 
        // Function call
        Console.Write("Position = "
                      + posOfRightMostDiffBit(m, n));
    }
}
 
// This code is contributed by shivanisinghss2110


PHP




<?php
// PHP implementation to find the
// position of rightmost different
// bit in two number.
 
// function to find rightmost
// different bit in two numbers.
function posOfRightMostDiffBit($m, $n)
{
    $t = floor(log($m ^ $n, 2));
    return $t;
}
 
// Driver code
$m = 52;
$n = 4;
 
// Function call
echo "Position = " ,
    posOfRightMostDiffBit($m, $n);
 
// This code is contributed by ajit
?>


Javascript




<script>
 
    // Javascript implementation to find the
    // position of rightmost different
    // bit in two number.
     
    // function to find rightmost
    // different bit in two numbers.
    function posOfRightMostDiffBit(m, n)
    {
      return parseInt(Math.floor
      (Math.log10(Math.pow(m ^ n, 2))), 10) + 2;
    }
     
    let m = 52, n = 4;
    document.write(
    "Position = " + posOfRightMostDiffBit(m, n)
    );
     
</script>


Output

Position = 5

Time Complexity: O(log2 N), this time complexity is equal to O(1) as one has to check for at most 31 bits only
Auxiliary Space: O(1)

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 review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or if you want to share more information about the topic discussed above.
 


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!