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 = 3Input: a = 5, b = 11, N = 1000001
Output: 14
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 <bits/stdc++.h>using namespace std;// Function to return the nth XOR Fibonacci numberint 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 codeint 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 = 1b = 2n = 10print(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 |
Javascript
<script>// Javascript implementation of the approach// Function to return the nth XOR Fibonacci numberfunction 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 let a = 1, b = 2, n = 10; document.write(nthXorFib(n, a, b));</script> |
2
Time Complexity: O(1)
Auxiliary Space: O(1)
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