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 Nth 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 Nth term would be same as the (N % 6)th term.

Below is the implementation of the above approach:

C++

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// 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;
}

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Java

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// 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

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Python3

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# 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] =
    f[1] =
  
    # 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

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C#

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// 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

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PHP

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<?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.
?>

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Output:

-2

Time Complexity: O(1)



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