Position of rightmost different bit - GeeksforGeeks

01 Jul, 2021
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 that 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. 
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++
Java
Python3
C#
PHP
Javascript
// 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 programint main(){    int m = 52, n = 24;       cout << "Position of rightmost different bit:"         << posOfRightMostDiffBit(m, n)<<endl;    return 0;}
// 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.
# 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 = 52n = 4print("position = ", int(posOfRightMostDiffBit(m, n))) # This code is contributed# by Anant Agarwal.
// C# implementation to find the position// of rightmost different bitusing 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.
<?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?>
<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>
Position of rightmost different bit:3
Using ffs() function
C++
Java
Python3
C#
PHP
Javascript
// 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 codeint main(){    int m = 52, n = 4;    cout <<"Position = " <<         posOfRightMostDiffBit(m, n);    return 0;}
// 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 codepublic static void main(String[] args){  int m = 52, n = 4;  System.out.println("Position = " +                      posOfRightMostDiffBit(m, n));}} // This code is contributed by gauravrajput1
# 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 codeif __name__ == '__main__':         m, n = 52, 4         print("Position = ",    posOfRightMostDiffBit(m, n)) # This code is contributed by mohit kumar 29
// 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;    Console.Write("Position = " +                  posOfRightMostDiffBit(m, n));  }} // This code is contributed by shivanisinghss2110
<?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?>
<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>
Position = 5
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