Given integers ‘K’ and ‘N’, the task is to find the Nth term of the K-Fibonacci series.
In K – Fibonacci series, the first ‘K’ terms will be ‘1’ and after that every ith term of the series will be the sum of previous ‘K’ elements in the same series.
Examples:
Input: N = 4, K = 2 Output: 3 The K-Fibonacci series for K=2 is 1, 1, 2, 3, ... And, the 4th element is 3. Input: N = 5, K = 6 Output: 1 The K-Fibonacci series for K=6 is 1, 1, 1, 1, 1, 1, 6, 11, ...
A simple approach:
- First, initialize the first ‘K’ elements to ‘1’.
- Then, calculate the sum of previous ‘K’ elements by running a loop from ‘i-k’ to ‘i-1’.
- Set the ith value to the sum.
Time Complexity: O(N*K)
An efficient approach:
- First, initialize the first ‘K’ elements to ‘1’.
- Create a variable named ‘sum’ which will be initialized with ‘K’.
- Set the value of (K+1)th element to sum.
- Set the next values as Array[i] = sum – Array[i-k-1] + Array[i-1] then update sum = Array[i].
- In the end, display the Nth term of the array.
Below is the implementation of the above approach:
C++
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std;// Function that finds the Nth// element of K-Fibonacci seriesvoid solve(int N, int K){ vector<long long int> Array(N + 1, 0); // If N is less than K // then the element is '1' if (N <= K) { cout << "1" << endl; return; } long long int i = 0, sum = K; // first k elements are 1 for (i = 1; i <= K; ++i) { Array[i] = 1; } // (K+1)th element is K Array[i] = sum; // find the elements of the // K-Fibonacci series for (int i = K + 2; i <= N; ++i) { // subtract the element at index i-k-1 // and add the element at index i-i // from the sum (sum contains the sum // of previous 'K' elements ) Array[i] = sum - Array[i - K - 1] + Array[i - 1]; // set the new sum sum = Array[i]; } cout << Array[N] << endl;}// Driver codeint main(){ long long int N = 4, K = 2; // get the Nth value // of K-Fibonacci series solve(N, K); return 0;} |
Java
// Java implementation of above approachpublic class GFG { // Function that finds the Nth // element of K-Fibonacci series static void solve(int N, int K) { int Array[] = new int[N + 1]; // If N is less than K // then the element is '1' if (N <= K) { System.out.println("1") ; return; } int i = 0 ; int sum = K; // first k elements are 1 for (i = 1; i <= K; ++i) { Array[i] = 1; } // (K+1)th element is K Array[i] = sum; // find the elements of the // K-Fibonacci series for (i = K + 2; i <= N; ++i) { // subtract the element at index i-k-1 // and add the element at index i-i // from the sum (sum contains the sum // of previous 'K' elements ) Array[i] = sum - Array[i - K - 1] + Array[i - 1]; // set the new sum sum = Array[i]; } System.out.println(Array[N]); } public static void main(String args[]) { int N = 4, K = 2; // get the Nth value // of K-Fibonacci series solve(N, K); } // This code is contributed by ANKITRAI1} |
Python3
# Python3 implementation of above approach# Function that finds the Nth# element of K-Fibonacci seriesdef solve(N, K) : Array = [0] * (N + 1) # If N is less than K # then the element is '1' if (N <= K) : print("1") return i = 0 sm = K # first k elements are 1 for i in range(1, K + 1) : Array[i] = 1 # (K+1)th element is K Array[i + 1] = sm # find the elements of the # K-Fibonacci series for i in range(K + 2, N + 1) : # subtract the element at index i-k-1 # and add the element at index i-i # from the sum (sum contains the sum # of previous 'K' elements ) Array[i] = sm - Array[i - K - 1] + Array[i - 1] # set the new sum sm = Array[i] print(Array[N]) # Driver codeN = 4K = 2# get the Nth value# of K-Fibonacci seriessolve(N, K)# This code is contributed by Nikita Tiwari. |
C#
// C# implementation of above approach using System;class GFG { // Function that finds the Nth // element of K-Fibonacci series public static void solve(int N, int K) { int[] Array = new int[N + 1]; // If N is less than K // then the element is '1' if (N <= K) { Console.WriteLine("1"); return; } int i = 0; int sum = K; // first k elements are 1 for (i = 1; i <= K; ++i) { Array[i] = 1; } // (K+1)th element is K Array[i] = sum; // find the elements of the // K-Fibonacci series for (i = K + 2; i <= N; ++i) { // subtract the element at index i-k-1 // and add the element at index i-i // from the sum (sum contains the sum // of previous 'K' elements ) Array[i] = sum - Array[i - K - 1] + Array[i - 1]; // set the new sum sum = Array[i]; } Console.WriteLine(Array[N]); } // Main Method public static void Main(string[] args) { int N = 4, K = 2; // get the Nth value // of K-Fibonacci series solve(N, K); } }// This code is contributed// by Shrikant13 |
PHP
<?php // PHP implementation of above approach// Function that finds the Nth// element of K-Fibonacci seriesfunction solve($N, $K){ $Array = array_fill(0, $N + 1, NULL); // If N is less than K // then the element is '1' if ($N <= $K) { echo "1" ."\n"; return; } $i = 0; $sum = $K; // first k elements are 1 for ($i = 1; $i <= $K; ++$i) { $Array[$i] = 1; } // (K+1)th element is K $Array[$i] = $sum; // find the elements of the // K-Fibonacci series for ($i = $K + 2; $i <= $N; ++$i) { // subtract the element at index i-k-1 // and add the element at index i-i // from the sum (sum contains the sum // of previous 'K' elements ) $Array[$i] = $sum - $Array[$i - $K - 1] + $Array[$i - 1]; // set the new sum $sum = $Array[$i]; } echo $Array[$N] . "\n";}// Driver code$N = 4;$K = 2;// get the Nth value// of K-Fibonacci seriessolve($N, $K);// This code is contributed // by ChitraNayal?> |
Javascript
<script>//Javascript program to find// next greater number than N // Function that finds the Nth// element of K-Fibonacci seriesfunction solve(N, K){ var Arr = new Array(N + 1); // If N is less than K // then the element is '1' if (N <= K) { document.write( "1" + "<br>"); return; } var i = 0, sum = K; // first k elements are 1 for (i = 1; i <= K; ++i) { Arr[i] = 1; } // (K+1)th element is K Arr[i] = sum; // find the elements of the // K-Fibonacci series for (var i = K + 2; i <= N; ++i) { // subtract the element at index i-k-1 // and add the element at index i-i // from the sum (sum contains the sum // of previous 'K' elements ) Arr[i] = sum - Arr[i - K - 1] + Arr[i - 1]; // set the new sum sum = Arr[i]; } document.write( Arr[N] + "<br>");}var N = 4, K = 2; // get the Nth value // of K-Fibonacci series solve(N, K); // This code is contributed by SoumikMondal</script> |
Output:
3
Time Complexity: O(N)
Auxiliary Space: O(N)
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