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

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, else, answer = -1

## C++

`// CPP to find n such that XOR of n and n+1 ` `// is equals to given n ` `#include <bits/stdc++.h> ` `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; ` `} ` |

*chevron_right*

*filter_none*

## 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. ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

## 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. ` |

*chevron_right*

*filter_none*

## PHP

`<?php ` `// PHP to find n such ` `// that XOR of n and n+1 ` `// is equals to given n ` ` ` `// function to return ` `// the required n ` `function` `xorCalc(` `$k` `) ` `{ ` ` ` `if` `(` `$k` `== 1) ` ` ` `return` `2; ` ` ` ` ` `// if k is of form 2^i-1 ` ` ` `if` `(((` `$k` `+ 1) & ` `$k` `) == 0) ` ` ` `return` `floor` `(` `$k` `/ 2); ` ` ` ` ` `return` `1; ` `} ` ` ` `// Driver Code ` `$k` `= 31; ` `echo` `xorCalc(` `$k` `); ` ` ` `// This code is contributed by vt_m. ` `?> ` |

*chevron_right*

*filter_none*

Output:

15

This article is contributed by **Shivam Pradhan (anuj_charm)**. 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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.

## Recommended Posts:

- Smallest number greater or equals to N such that it has no odd positioned bit set
- Find the number of pairs such that their gcd is equals to 1
- Find smallest number formed by inverting digits of given number N
- Find smallest possible Number from a given large Number with same count of digits
- Find the smallest number whose digits multiply to a given number n
- Find smallest number K such that K % p = 0 and q % K = 0
- Find K'th smallest number such that A + B = A | B
- Given a number, find the next smallest palindrome
- Find the kth smallest number with sum of digits as m
- Find smallest permutation of given number
- Find the Smallest number that divides X^X
- Find the smallest number whose sum of digits is N
- Find the k-th smallest divisor of a natural number N
- Find the smallest number X such that X! contains at least Y trailing zeros.
- Find Nth smallest number that is divisible by 100 exactly K times
- Find smallest positive number Y such that Bitwise AND of X and Y is Zero
- Find the smallest number with n set and m unset bits
- Find the smallest positive number which can not be represented by given digits
- Find kth smallest number in range [1, n] when all the odd numbers are deleted
- Find the smallest positive number missing from an unsorted array | Set 3