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Position of rightmost common bit in two numbers
  • Difficulty Level : Easy
  • Last Updated : 12 Apr, 2021

Given two non-negative numbers m and n. Find the position of rightmost same bit in the binary representation of the numbers.

Examples:  

Input : m = 10, n = 9
Output : 3
(10)10 = (1010)2
(9)10 = (1001)2
It can be seen that the 3rd bit
from the right is same.

Input : m = 16, n = 7
Output : 4
(16)10 = (10000)2
(7)10 = (111)2, can also be written as
     = (00111)2
It can be seen that the 4th bit
from the right is same.

Approach: Get the bitwise xor of m and n. Let it be xor_value = m ^ n. Now, get the position of rightmost unset bit in xor_value.

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

C++




// C++ implementation to find the position
// of rightmost same bit
#include <bits/stdc++.h>
 
using namespace std;
 
// Function to find the position of
// rightmost set bit in 'n'
int getRightMostSetBit(unsigned int n)
{
    return log2(n & -n) + 1;
}
 
// Function to find the position of
// rightmost same bit in the
// binary representations of 'm' and 'n'
int posOfRightMostSameBit(unsigned int m,
                          unsigned int n)
{
    // position of rightmost same bit
    return getRightMostSetBit(~(m ^ n));
}
 
// Driver program to test above
int main()
{
    int m = 16, n = 7;
    cout << "Position = "
         << posOfRightMostSameBit(m, n);
    return 0;
}

Java




// Java implementation to find the position
// of rightmost same bit
class GFG {
         
    // Function to find the position of
    // rightmost set bit in 'n'
    static int getRightMostSetBit(int n)
    {
        return (int)((Math.log(n & -n))/(Math.log(2)))
                                                  + 1;
    }
     
    // Function to find the position of
    // rightmost same bit in the
    // binary representations of 'm' and 'n'
    static int posOfRightMostSameBit(int m,int n)
    {
         
        // position of rightmost same bit
        return getRightMostSetBit(~(m ^ n));
    }
     
    //Driver code
    public static void main (String[] args)
    {
        int m = 16, n = 7;
         
        System.out.print("Position = "
            + posOfRightMostSameBit(m, n));
    }
}
 
// This code is contributed by Anant Agarwal.

Python3




# Python3 implementation to find the
# position of rightmost same bit
import math
 
# Function to find the position
# of rightmost set bit in 'n'
def getRightMostSetBit(n):
 
    return int(math.log2(n & -n)) + 1
 
# Function to find the position of
# rightmost same bit in the binary
# representations of 'm' and 'n'
def posOfRightMostSameBit(m, n):
 
    # position of rightmost same bit
    return getRightMostSetBit(~(m ^ n))
 
# Driver Code
m, n = 16, 7
print("Position = ", posOfRightMostSameBit(m, n))
 
# This code is contributed by Anant Agarwal.

C#




// C# implementation to find the position
// of rightmost same bit
using System;
 
class GFG
{
    // Function to find the position of
    // rightmost set bit in 'n'
    static int getRightMostSetBit(int n)
    {
        return (int)((Math.Log(n & -n)) / (Math.Log(2))) + 1;
    }
      
    // Function to find the position of
    // rightmost same bit in the
    // binary representations of 'm' and 'n'
    static int posOfRightMostSameBit(int m,int n)
    {
        // position of rightmost same bit
        return getRightMostSetBit(~(m ^ n));
    }
     
    //Driver code
    public static void Main ()
    {
        int m = 16, n = 7;
        Console.Write("Position = "
              + posOfRightMostSameBit(m, n));
    }
}
//This code is contributed by Anant Agarwal.

PHP




<?php
// PHP implementation to
// find the position
// of rightmost same bit
 
// Function to find the position of
// rightmost set bit in 'n'
function getRightMostSetBit($n)
{
    return log($n & -$n) + 1;
}
 
// Function to find the position of
// rightmost same bit in the
// binary representations of 'm' and 'n'
function posOfRightMostSameBit($m,
                               $n)
{
     
    // position of rightmost same bit
    return getRightMostSetBit(~($m ^ $n));
}
 
    // Driver Code
    $m = 16; $n = 7;
    echo "Position = "
        , ceil(posOfRightMostSameBit($m, $n));
         
// This code is contributed by anuj_67.
?>

Javascript




<script>
// JavaScript implementation to find the position
// of rightmost same bit
 
// Function to find the position of
// rightmost set bit in 'n'
function getRightMostSetBit(n)
{
    return Math.log2(n & -n) + 1;
}
 
// Function to find the position of
// rightmost same bit in the
// binary representations of 'm' and 'n'
function posOfRightMostSameBit(m, n)
{
 
    // position of rightmost same bit
    return getRightMostSetBit(~(m ^ n));
}
 
// Driver program to test above
    let m = 16, n = 7;
    document.write("Position = "
        + posOfRightMostSameBit(m, n));
 
// This code is contributed by Manoj.
</script>

Output: 



Position = 4

Alternate Approach: Until both the value becomes zero, check last bits of both numbers and right shift. At any moment, both bits are same, return counter.

Explanation: Rightmost bit of two values m and n are equal only when both values are either odd or even. 

C++




// C++ implementation to find the position
// of rightmost same bit
#include <iostream>
using namespace std;
     
// Function to find the position of
// rightmost same bit in the binary
// representations of 'm' and 'n'
static int posOfRightMostSameBit(int m, int n)
{
     
    // Initialize loop counter
    int loopCounter = 1;
     
    while (m > 0 || n > 0)
    {
         
        // Check whether the value 'm' is odd
        bool a = m % 2 == 1;
         
        // Check whether the value 'n' is odd
        bool b = n % 2 == 1;
         
        // Below 'if' checks for both
        // values to be odd or even
        if (!(a ^ b))
        {
            return loopCounter;
        }
         
        // Right shift value of m
        m = m >> 1;
         
        // Right shift value of n
        n = n >> 1;
        loopCounter++;
    }
     
    // When no common set is found
    return -1;
}
 
// Driver code
int main()
{
    int m = 16, n = 7;
     
    cout << "Position = "
         <<  posOfRightMostSameBit(m, n);
}
 
// This code is contributed by shivanisinghss2110

Java




// Java implementation to find the position
// of rightmost same bit
class GFG {
     
    // Function to find the position of
    // rightmost same bit in the
    // binary representations of 'm' and 'n'
    static int posOfRightMostSameBit(int m,int n)
    {
        int loopCounter = 1; // Initialize loop counter
        while (m > 0 || n > 0){
             
            boolean a = m%2 == 1; //Check whether the value 'm' is odd
            boolean b = n%2 == 1; //Check whether the value 'n' is odd
             
            // Below 'if' checks for both values to be odd or even
            if (!(a ^ b)){
                return loopCounter;}
             
            m = m >> 1; //Right shift value of m
            n = n >> 1; //Right shift value of n
            loopCounter++;
        }
        return -1; //When no common set is found
    }
       
    //Driver code
    public static void main (String[] args)
    {
        int m = 16, n = 7;
           
        System.out.print("Position = "
            + posOfRightMostSameBit(m, n));
    }
}

Python3




# Python3 implementation to find the position
# of rightmost same bit
 
# Function to find the position of
# rightmost same bit in the
# binary representations of 'm' and 'n'
def posOfRightMostSameBit(m, n):
     
    # Initialize loop counter
    loopCounter = 1
     
    while (m > 0 or n > 0):
         
        # Check whether the value 'm' is odd
        a = m % 2 == 1
         
        # Check whether the value 'n' is odd
        b = n % 2 == 1
 
        # Below 'if' checks for both
        # values to be odd or even
        if (not (a ^ b)):
            return loopCounter
             
        # Right shift value of m
        m = m >> 1
         
        # Right shift value of n
        n = n >> 1
        loopCounter += 1
         
    # When no common set is found
    return -1
 
# Driver code
if __name__ == '__main__':
     
    m, n = 16, 7
 
    print("Position = ",
    posOfRightMostSameBit(m, n))
 
# This code is contributed by mohit kumar 29

C#




// C# implementation to find the position
// of rightmost same bit
using System;
class GFG
{
 
  // Function to find the position of
  // rightmost same bit in the
  // binary representations of 'm' and 'n'
  static int posOfRightMostSameBit(int m, int n)
  {
    int loopCounter = 1; // Initialize loop counter
    while (m > 0 || n > 0)
    {
 
      Boolean a = m % 2 == 1; // Check whether the value 'm' is odd
      Boolean b = n % 2 == 1; // Check whether the value 'n' is odd
 
      // Below 'if' checks for both values to be odd or even
      if (!(a ^ b))
      {
        return loopCounter;
      }
 
      m = m >> 1; // Right shift value of m
      n = n >> 1; // Right shift value of n
      loopCounter++;
    }
    return -1; // When no common set is found
  }
 
  // Driver code
  public static void Main (String[] args)
  {
    int m = 16, n = 7;        
    Console.Write("Position = "
                  + posOfRightMostSameBit(m, n));
  }
}
 
// This code is contributed by shivanisinghss2110

Javascript




<script>
// JavaScript implementation to find the position
// of rightmost same bit
     
// Function to find the position of
// rightmost same bit in the binary
// representations of 'm' and 'n'
function posOfRightMostSameBit(m, n)
{
     
    // Initialize loop counter
    let loopCounter = 1;
     
    while (m > 0 || n > 0)
    {
         
        // Check whether the value 'm' is odd
        let a = m % 2 == 1;
         
        // Check whether the value 'n' is odd
        let b = n % 2 == 1;
         
        // Below 'if' checks for both
        // values to be odd or even
        if (!(a ^ b))
        {
            return loopCounter;
        }
         
        // Right shift value of m
        m = m >> 1;
         
        // Right shift value of n
        n = n >> 1;
        loopCounter++;
    }
     
    // When no common set is found
    return -1;
}
 
// Driver code
    let m = 16, n = 7;
     
    document.write("Position = "
        + posOfRightMostSameBit(m, n));
 
// This code is contributed by Manoj.
</script>

Output: 

Position = 4

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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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