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Position of rightmost different bit
• Difficulty Level : Easy
• Last Updated : 12 Jun, 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++

 `// C++ implementation to find the position``// of rightmost different bit``#include ``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 program``int` `main()``{``    ``int` `m = 52, n = 24;``  ` `    ``cout << ``"Position of rightmost different bit:"``         ``<< posOfRightMostDiffBit(m, n)<

## 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``;``        ``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` `m ``=` `52``n ``=` `4``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;``        ``Console.Write(``"Position = "` `+``            ``posOfRightMostDiffBit(m, n));``    ``}``}` `// This code is contributed by Smitha.`

## PHP

 ``

## Javascript

 ``
Output
`Position of rightmost diffrent bit:3`

Using ffs() function

## C++

 `// C++ implementation to find the``// position of rightmost different``// bit in two number.``#include ``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;``}`

## 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``;``  ``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``    ` `    ``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;``    ``Console.Write(``"Position = "` `+``                  ``posOfRightMostDiffBit(m, n));``  ``}``}` `// This code is contributed by shivanisinghss2110`

## PHP

 ``

## Javascript

 ``
Output
`Position = 5`

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