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

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; ` `} ` |

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## 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. ` |

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## 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 ` |

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## 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. ` |

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## 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. ` `?> ` |

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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.

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