Related Articles
Nth XOR Fibonacci number
• Difficulty Level : Medium
• Last Updated : 09 Sep, 2019

Given three integers a, b and N where a and b are the first two terms of the XOR Fibonacci series and the task is to find the Nth term.
The Nth term of the XOR Fibonacci series is defined as F(N) = F(N – 1) ^ F(N – 2) where ^ is the bitwise XOR.

Examples:

Input: a = 1, b = 2, N = 5
Output: 3
F(0) = 1
F(1) = 2
F(2) = 1 ^ 2 = 3
F(3) = 2 ^ 3 = 1
F(4) = 1 ^ 3 = 2
F(5) = 1 ^ 2 = 3

Input: a = 5, b = 11, N = 1000001
Output: 14

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Since, a ^ a = 0 and it is given that

F(0) = a and F(1) = b
Now, F(2) = F(0) ^ F(1) = a ^ b
And, F(3) = F(1) ^ F(2) = b ^ (a ^ b) = a
F(4) = a ^ b ^ a = b
F(5) = a ^ b
F(6) = a
F(7) = b
F(8) = a ^ b

It can be observed that the answer repeats itself after every 3 numbers. So the answer is F(N % 3) where F(0) = a, F(1) = b and F(2) = a ^ b.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;`` ` `// Function to return the nth XOR Fibonacci number``int` `nthXorFib(``int` `n, ``int` `a, ``int` `b)``{``    ``if` `(n == 0)``        ``return` `a;``    ``if` `(n == 1)``        ``return` `b;``    ``if` `(n == 2)``        ``return` `(a ^ b);`` ` `    ``return` `nthXorFib(n % 3, a, b);``}`` ` `// Driver code``int` `main()``{``    ``int` `a = 1, b = 2, n = 10;`` ` `    ``cout << nthXorFib(n, a, b);`` ` `    ``return` `0;``}`

## Java

 `// Java implementation of the above approach ``class` `GFG``{``         ` `    ``// Function to return the ``    ``// nth XOR Fibonacci number ``    ``static` `int` `nthXorFib(``int` `n, ``int` `a, ``int` `b) ``    ``{ ``        ``if` `(n == ``0``) ``            ``return` `a; ``        ``if` `(n == ``1``) ``            ``return` `b; ``        ``if` `(n == ``2``) ``            ``return` `(a ^ b); ``     ` `        ``return` `nthXorFib(n % ``3``, a, b); ``    ``} ``     ` `    ``// Driver code ``    ``public` `static` `void` `main (String[] args) ``    ``{ ``        ``int` `a = ``1``, b = ``2``, n = ``10``; ``     ` `        ``System.out.println(nthXorFib(n, a, b)); ``    ``} ``}`` ` `// This code is contributed by AnkitRai01`

## Python3

 `# Python3 implementation of the approach `` ` `# Function to return``# the nth XOR Fibonacci number ``def` `nthXorFib(n, a, b):``    ``if` `n ``=``=` `0` `: ``        ``return` `a ``    ``if` `n ``=``=` `1` `: ``        ``return` `b ``    ``if` `n ``=``=` `2` `: ``        ``return` `a ^ b `` ` `    ``return` `nthXorFib(n ``%` `3``, a, b) `` ` `# Driver code ``a ``=` `1``b ``=` `2``n ``=` `10``print``(nthXorFib(n, a, b)) `` ` `# This code is contributed by divyamohan123 `

## C#

 `// C# implementation of the above approach ``using` `System;``     ` `class` `GFG``{``         ` `    ``// Function to return the ``    ``// nth XOR Fibonacci number ``    ``static` `int` `nthXorFib(``int` `n, ``int` `a, ``int` `b) ``    ``{ ``        ``if` `(n == 0) ``            ``return` `a; ``        ``if` `(n == 1) ``            ``return` `b; ``        ``if` `(n == 2) ``            ``return` `(a ^ b); ``     ` `        ``return` `nthXorFib(n % 3, a, b); ``    ``} ``     ` `    ``// Driver code ``    ``public` `static` `void` `Main (String[] args) ``    ``{ ``        ``int` `a = 1, b = 2, n = 10; ``     ` `        ``Console.WriteLine(nthXorFib(n, a, b)); ``    ``} ``}`` ` `// This code is contributed by Princi Singh`
Output:
```2
```

Time Complexity: O(1) My Personal Notes arrow_drop_up