# 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++

 `// 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 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)<

## 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

 ``

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`

## PHP

 ``

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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