Find the Nth term of the series where each term f[i] = f[i – 1] – f[i – 2]

Given three integers X, Y and N, the task is to find the N^th term of the series f[i] = f[i – 1] – f[i – 2], i > 1 where f[0] = X and f[1] = Y.
Examples:

Input: X = 2, Y = 3, N = 3 
Output: -2 
The series will be 2 3 1 -2 -3 -1 2 and f[3] = -2
Input: X = 3, Y = 7, N = 8 
Output: 4 

Approach: An important observation here is that there will be atmost 6 distinct terms, before the sequence starts repeating itself. So, find the first 6 terms of the series and then the N^th term would be same as the (N % 6)^th term.
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 term
// of the given series
int findNthTerm(int x, int y, int n)
{
    int f[6];

    // First and second term of the series
    f[0] = x;
    f[1] = y;

    // Find first 6 terms
    for (int i = 2; i <= 5; i++)
        f[i] = f[i - 1] - f[i - 2];

    // Return the Nth term
    return f[n % 6];
}

// Driver code
int main()
{
    int x = 2, y = 3, n = 3;
    cout << findNthTerm(x, y, n);

    return 0;
}

Java

// Java implementation of the approach
import java.io.*;

class GFG
{

    // Function to find the nth term of series
    static int findNthTerm(int x, int y, int n)
    {
        int[] f = new int[6];

        f[0] = x;
        f[1] = y;

        // Loop to add numbers
        for (int i = 2; i <= 5; i++)
            f[i] = f[i - 1] - f[i - 2];

        return f[n % 6];
    }


    // Driver code
    public static void main(String args[])
    {
        int x = 2, y = 3, n = 3;
        System.out.println(findNthTerm(x, y, n));
    }
}

// This code is contributed by mohit kumar 29

Python3

# Python3 implementation of the approach

# Function to return the Nth term
# of the given series
def findNthTerm(x, y, n):

    f = [0] * 6

    # First and second term of
    # the series
    f[0] = x
    f[1] = y

    # Find first 6 terms
    for i in range(2, 6):
        f[i] = f[i - 1] - f[i - 2]

    # Return the Nth term
    return f[n % 6]

# Driver code
if __name__ == "__main__":

    x, y, n = 2, 3, 3
    print(findNthTerm(x, y, n))

# This code is contributed by
# Rituraj Jain

C#

// C# implementation of the approach
using System;

class GFG
{

    // Function to find the nth term of series
    static int findNthTerm(int x, int y, int n)
    {
        int[] f = new int[6];

        f[0] = x;
        f[1] = y;

        // Loop to add numbers
        for (int i = 2; i <= 5; i++)
            f[i] = f[i - 1] - f[i - 2];

        return f[n % 6];
    }

    // Driver code
    public static void Main()
    {
        int x = 2, y = 3, n = 3;
        Console.WriteLine(findNthTerm(x, y, n));
    }
}

// This code is contributed by Ryuga

PHP

<?php

//PHP implementation of the approach
// Function to find the nth term of series
function findNthTerm($x, $y, $n)
{
    $f = array(6);

    $f[0] = $x;
    $f[1] = $y;

    // Loop to add numbers
    for ($i = 2; $i <= 5; $i++)
        $f[$i] = $f[$i - 1] - $f[$i - 2];

    return $f[$n % 6];
}

// Driver code
$x = 2; $y = 3; $n = 3;
echo(findNthTerm($x, $y, $n));

// This code is contributed by Code_Mech.
?>

Javascript

<script>

// JavaScript implementation of the approach

    // Function to return the Nth term
    // of the given series
    function findNthTerm(x, y, n)
    {
        let f = new Array(6);

        // First and second term of the series
        f[0] = x;
        f[1] = y;

        // Find first 6 terms
        for (let i = 2; i <= 5; i++)
            f[i] = f[i - 1] - f[i - 2];

        // Return the Nth term
        return f[n % 6];
    }

    // Driver code

    let x = 2, y = 3, n = 3;
    document.write(findNthTerm(x, y, n));

// This code is contributed by Surbhi Tyagi

</script>

Output:

-2

Time Complexity: O(1)