Position of rightmost different bit

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
(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
(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            

Approach: 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.

Explanation: The bitwise xor operation produces a number which has set bits only at the positions where the bits of m and n differ. Thus, the position of the rightmost set bit in xor_value gives the position of the rightmost different bit. 

Below is the implementation of the above approach:
 



C++

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// 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 diffrent bit.
int posOfRightMostDiffBit(int m, int n)
{
    // position of rightmost different
    //  bit
     
    return getRightMostSetBit(m ^ n);
}
 
// Driver program
int main()
{
    int m = 52, n = 24;
   
    cout << "Position of rightmost diffrent bit:"
         << posOfRightMostDiffBit(m, n)<<endl;
    return 0;
}

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Java

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// 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;
        System.out.print("Position = " +
            posOfRightMostDiffBit(m, n));
    }
}
 
// This code is contributed by Anant Agarwal.

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Python3

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# 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
 
m = 52
n = 4
print("position = ", int(posOfRightMostDiffBit(m, n)))
 
# This code is contributed
# by Anant Agarwal.

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C#

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// 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;
        Console.Write("Position = " +
            posOfRightMostDiffBit(m, n));
    }
}
 
// This code is contributed by Smitha.

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PHP

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<?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;
    $n = 4;
    echo posOfRightMostDiffBit($m, $n);
         
// This code is contributed by Ajit
?>

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Output

Position of rightmost diffrent bit:3



Using ffs() function

C++

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// 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;
    cout <<"Position = " <<
         posOfRightMostDiffBit(m, n);
    return 0;
}

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Java

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// 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;
  System.out.println("Position = "
                     posOfRightMostDiffBit(m, n));
}
}
 
// This code is contributed by gauravrajput1

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PHP

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<?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;
$n = 4;
echo "Position = " ,
    posOfRightMostDiffBit($m, $n);
 
// This code is contributed by ajit
?>

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Output

Position = 5


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