# Find smallest number n such that n XOR n+1 equals to given k.

• Difficulty Level : Hard
• Last Updated : 06 Oct, 2021

You are given a positive number k, we need to find a positive integer n, such that XOR of n and n+1 is equal to k. If no such n exist then print -1.
Examples:

```Input : 3
Output : 1

Input : 7
Output : 3

Input : 6
Output : -1```

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Below are two cases when we do n XOR (n+1) for a number n.
Case 1 : n is even. Last bit of n is 0 and last bit of (n+1) is 1. Rest of the bits are same in both. So XOR would always be 1 if n is even.
Case : n is odd Last bit in n is 1. And in n+1, last bit is 0. But in this case there may be more bits which differ due to carry. The carry continues to propagate to left till we find first 0 bit. So n XOR n+1 will we 2^i-1 where i is the position of first 0 bit in n from left. So, we can say that if k is of form 2^i-1 then we will have our answer as k/2.
Finally our steps are:

```If we have k=1, answer = 2 [We need smallest positive n]
Else If k is of form 2^i-1, answer = k/2,

## C++

 `// CPP to find n such that XOR of n and n+1``// is equals to given n``#include ``using` `namespace` `std;` `// function to return the required n``int` `xorCalc(``int` `k)``{``    ``if` `(k == 1)``        ``return` `2;``    ` `    ``// if k is of form 2^i-1``    ``if` `(((k + 1) & k) == 0)``        ``return` `k / 2;` `    ``return` `-1;``}` `// driver program``int` `main()``{``    ``int` `k = 31;``    ``cout << xorCalc(k);``    ``return` `0;``}`

## Java

 `// Java to find n such that XOR of n and n+1``// is equals to given n``class` `GFG``{``    ` `    ``// function to return the required n``    ``static` `int` `xorCalc(``int` `k)``    ``{``        ``if` `(k == ``1``)``            ``return` `2``;``        ` `        ``// if k is of form 2^i-1``        ``if` `(((k + ``1``) & k) == ``0``)``            ``return` `k / ``2``;``    ` `        ``return` `1``;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `k = ``31``;``        ` `        ``System.out.println(xorCalc(k));``    ``}``}` `// This code is contributed by Anant Agarwal.`

## Python3

 `# python to find n such that``# XOR of n and n+1 is equals``# to given n` `# function to return the``# required n``def` `xorCalc(k):``    ``if` `(k ``=``=` `1``):``        ``return` `2``    ` `    ``# if k is of form 2^i-1``    ``if` `(((k ``+` `1``) & k) ``=``=` `0``):``        ``return` `k ``/` `2` `    ``return` `1``;`  `# driver program``k ``=` `31``print``(``int``(xorCalc(k)))` `# This code is contributed``# by Sam007`

## C#

 `// C# to find n such that XOR``// of n and n+1 is equals to``// given n``using` `System;` `class` `GFG``{``    ` `    ``// function to return the required``    ``// n``    ``static` `int` `xorCalc(``int` `k)``    ``{``        ``if` `(k == 1)``            ``return` `2;``        ` `        ``// if k is of form 2^i-1``        ``if` `(((k + 1) & k) == 0)``            ``return` `k / 2;``    ` `        ``return` `1;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main ()``    ``{``        ``int` `k = 31;``        ` `        ``Console.WriteLine(xorCalc(k));``    ``}``}` `// This code is contributed by vt_m.`

## PHP

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## Javascript

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Output:

`15`