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++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the Nth term// of the given seriesint 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 codeint main(){ int x = 2, y = 3, n = 3; cout << findNthTerm(x, y, n); return 0;} |
Java
// Java implementation of the approachimport 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 seriesdef 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 codeif __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 approachusing 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 seriesfunction 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> |
-2
Time Complexity: O(1)
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