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

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

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. 
?> 

Output:

-2

Time Complexity: O(1)

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My Personal Notes

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

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

C++

Java

Python3

C#

PHP

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