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Position of rightmost common bit in two numbers

Last Updated : 21 Jun, 2022
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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

Time Complexity: O(1)

Auxiliary Space: O(1)

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

Time Complexity: O(1)

Auxiliary Space: O(1)


 


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