N-bonacci Numbers
You are given two integers N and M, and print all the terms of the series up to M-terms of the N-bonacci Numbers. For example, when N = 2, the sequence becomes Fibonacci, when n = 3, sequence becomes Tribonacci.
In general, in N-bonacci sequence, we use sum of preceding N numbers from the next term. For example, a 3-bonacci sequence is the following:
0, 0, 1, 1, 2, 4, 7, 13, 24, 44, 81
The Fibonacci sequence is a set of numbers that starts with one or zero, followed by a one, and proceeds based on the rule that each number is equal to the sum of preceding two numbers 0, 1, 1, 2, 3, 5, 8…..
Examples :
Input : N = 3, M = 8
Output : 0, 0, 1, 1, 2, 4, 7, 13
We need to print first M terms.
First three terms are 0, 0 and 1.
Fourth term is 0 + 0 + 1 = 1
Fifth term is 0 + 1 + 1 = 2
Sixth terms is 1 + 1 + 2 = 4
Seventh term is 7 (1 + 2 + 4)
and eighth term is 13 (7 + 4 + 2).Input : N = 4, M = 10
Output : 0 0 0 1 1 2 4 8 15 29
Method 1 (Simple)
Initialize first N-1 terms as 0 and N-th term as 1. Now to find terms from (N+1)-th to M-th, we simply compute sum of previous N terms.
Example : N = 4, M = 10
First three terms are 0, 0, 0
Fourth term is 1.
Remaining terms are computed by adding
previous 4 terms.
0 0 0 1
0 0 0 1 1
0 0 0 1 1 2
0 0 0 1 1 2 4
0 0 0 1 1 2 4 8
0 0 0 1 1 2 4 8 15
0 0 0 1 1 2 4 8 15 29
C++
// CPP program print first M terms of// N-bonacci series.#include <bits/stdc++.h>using namespace std;// Function to print bonacci seriesvoid bonacciseries(long n, int m){ // Assuming m >= n. int a[m] = { 0 }; a[n - 1] = 1; // Computing every term as sum of previous // n terms. for (int i = n; i < m; i++) for (int j = i - n; j < i; j++) a[i] += a[j]; for (int i = 0; i < m; i++) cout << a[i] << " ";}// Driver's Codeint main(){ int N = 5, M = 15; bonacciseries(N, M); return 0;} |
Java
// Java program print first M// terms of N-bonacci series.import java.io.*;class GFG { // Function to print // bonacci series static void bonacciseries(int n, int m) { // Assuming m >= n. int[] a = new int[m]; a[n - 1] = 1; // Computing every term as // sum of previous n terms. for (int i = n; i < m; i++) for (int j = i - n; j < i; j++) a[i] += a[j]; for (int i = 0; i < m; i++) System.out.print(a[i] + " "); } // Driver Code public static void main(String args[]) { int N = 5, M = 15; bonacciseries(N, M); }}// This code is contributed by// Manish Shaw(manishshaw1) |
Python3
# Python program print first M# terms of N-bonacci series.# Function to print bonacci seriesdef bonacciseries(n, m): # Assuming m >= n. a = [0] * m a[n - 1] = 1 # Computing every term as # sum of previous n terms. for i in range(n, m): for j in range(i - n, i): a[i] = a[i] + a[j] for i in range(0, m): print(a[i], end=" ")# Driver CodeN = 5M = 15bonacciseries(N, M)# This code is contributed# by Manish Shaw(manishshaw1) |
C#
// C# program print first M// terms of N-bonacci series.using System;class GFG { // Function to print // bonacci series static void bonacciseries(int n, int m) { // Assuming m >= n. int[] a = new int[m]; Array.Clear(a, 0, a.Length); a[n - 1] = 1; // Computing every term as // sum of previous n terms. for (int i = n; i < m; i++) for (int j = i - n; j < i; j++) a[i] += a[j]; for (int i = 0; i < m; i++) Console.Write(a[i] + " "); } // Driver Code static void Main() { int N = 5, M = 15; bonacciseries(N, M); }}// This code is contributed by// Manish Shaw(manishshaw1) |
PHP
<?php// PHP program print first M // terms of N-bonacci series.// Function to print bonacci seriesfunction bonacciseries($n, $m){ // Assuming m >= n. $a = array_fill(0, $m, 0); $a[$n - 1] = 1; // Computing every term as // sum of previous n terms. for ($i = $n; $i < $m; $i++) for ($j = $i - $n; $j < $i; $j++) $a[$i] += $a[$j]; for ($i = 0; $i < $m; $i++) echo ($a[$i]." ");}// Driver Code$N = 5; $M = 15;bonacciseries($N, $M);// This code is contributed // by Manish Shaw(manishshaw1)?> |
Javascript
<script>// Javascript program print first M // terms of N-bonacci series. // Function to print // bonacci series function bonacciseries(n , m) { // Assuming m >= n. var a = Array(m).fill(0); a[n - 1] = 1; // Computing every term as // sum of previous n terms. for (i = n; i < m; i++) for (j = i - n; j < i; j++) a[i] += a[j]; for (i = 0; i < m; i++) document.write(a[i] + " "); } // Driver Code var N = 5, M = 15; bonacciseries(N, M);// This code contributed by Rajput-Ji </script> |
Output :
0 0 0 0 1 1 2 4 8 16 31 61 120 236 464
Time Complexity: O(M * N)
Auxiliary Space: O(M)
Method 2 (Optimized)
We can optimize for large values of N. The idea is based on sliding window. The current term a[i] can be computed as a[i-1] + a[i-1] – a[i-n-1]
C++
// CPP program print first M terms of// N-bonacci series.#include <bits/stdc++.h>using namespace std;// Function to print bonacci seriesvoid bonacciseries(long n, int m){ // Assuming m > n. int a[m] = { 0 }; a[n - 1] = 1; a[n] = 1; // Uses sliding window for (int i = n + 1; i < m; i++) a[i] = 2 * a[i - 1] - a[i - n - 1]; // Printing result for (int i = 0; i < m; i++) cout << a[i] << " ";}// Driver's Codeint main(){ int N = 5, M = 15; bonacciseries(N, M); return 0;} |
Java
// Java program print first M terms of// N-bonacci series.class GFG { // Function to print bonacci series static void bonacciseries(int n, int m) { // Assuming m > n. int a[] = new int[m]; for (int i = 0; i < m; i++) a[i] = 0; a[n - 1] = 1; a[n] = 1; // Uses sliding window for (int i = n + 1; i < m; i++) a[i] = 2 * a[i - 1] - a[i - n - 1]; // Printing result for (int i = 0; i < m; i++) System.out.print(a[i] + " "); } // Driver's Code public static void main(String args[]) { int N = 5, M = 15; bonacciseries(N, M); }}// This code is contributed by JaideepPyne. |
Python3
# Python3 program print first M terms of# N-bonacci series.# Function to print bonacci seriesdef bonacciseries(n, m): # Assuming m > n. a = [0 for i in range(m)] a[n - 1] = 1 a[n] = 1 # Uses sliding window for i in range(n + 1, m): a[i] = 2 * a[i - 1] - a[i - n - 1] # Printing result for i in range(0, m): print(a[i], end=" ")# Driver's Codeif __name__ == '__main__': N, M = 5, 15 bonacciseries(N, M)# This code is contributed by# Sanjit_Prasad |
C#
// Java program print// first M terms of// N-bonacci series.using System;class GFG { // Function to print // bonacci series static void bonacciseries(int n, int m) { // Assuming m > n. int[] a = new int[m]; for (int i = 0; i < m; i++) a[i] = 0; a[n - 1] = 1; a[n] = 1; // Uses sliding window for (int i = n + 1; i < m; i++) a[i] = 2 * a[i - 1] - a[i - n - 1]; // Printing result for (int i = 0; i < m; i++) Console.Write(a[i] + " "); } // Driver Code static void Main() { int N = 5, M = 15; bonacciseries(N, M); }}// This code is contributed by// Manish Shaw(manishshaw1) |
PHP
<?php// PHP program print // first M terms of// N-bonacci series.// Function to print // N-bonacci seriesfunction bonacciseries($n, $m){ // Assuming m > n. $a = array(); for ($i = 0; $i < $m; $i++) $a[$i] = 0; $a[$n - 1] = 1; $a[$n] = 1; // Uses sliding window for ($i = $n + 1; $i < $m; $i++) $a[$i] = 2 * $a[$i - 1] - $a[$i - $n - 1]; // Printing result for ($i = 0; $i < $m; $i++) echo ($a[$i] . " ");}// Driver Code$N = 5; $M = 15;bonacciseries($N, $M);// This code is contributed by // Manish Shaw(manishshaw1)?> |
Javascript
<script>// Javascript program print first M terms of// N-bonacci series.// Function to print bonacci series function bonacciseries(n , m) { // Assuming m > n. var a = Array(m).fill(0); for (i = 0; i < m; i++) a[i] = 0; a[n - 1] = 1; a[n] = 1; // Uses sliding window for (i = n + 1; i < m; i++) a[i] = 2 * a[i - 1] - a[i - n - 1]; // Printing result for (i = 0; i < m; i++) document.write(a[i] + " "); } // Driver's Code var N = 5, M = 15; bonacciseries(N, M);// This code contributed by Rajput-Ji </script> |
Output:
0 0 0 0 1 1 2 4 8 16 31 61 120 236 464
Time Complexity: O(M)
Auxiliary Space: O(M)
Method 3: Using Queue
We can use queue to easily and efficiently calculate every terms of N-bonacci numbers by using less space than above method.
Algorithm:
- First we push N element inside the queue whose values will be ‘0‘ for 1 to N-1 elements and ‘1‘ for Nth element.
- Now, we take another variable (let’s say su) to calculate N-bonacci numbers. and set its value to 1, which is the sum of all the elements, currently inside the queue.
- Then, using a for loop iterate the queue M times:
- Inside for loop, we pop the front element of the queue and store it.
- Then, we push an element whose value will be sum of all the elements, currently inside the queue.
- Push su into the queue.
- Before popping, the sum of all the elements is su, and then we push su inside queue.
- So, now the sum is 2*su.
- And, then we pop front element of the queue, so sum will be (2*su – popped_front_element).
- So, su will now become (2*su – popped_front_element), which we will be pushed inside the queue in next iteration.
- And then we print the popped element.
Below is the implementation of the above approach:
C++
// CPP program print first M terms of// N-bonacci series using queue#include <bits/stdc++.h>using namespace std;// Function to find and print N-bonacci seriesvoid bonacciseries(int n, long m){ queue<int> q; // Assuming m > n. for (int i = 0; i < (n - 1); i++) { q.push(0); } // Nth Element in the queue q.push(1); // Initilize value of su as the sum of all the elements, // currently inside the queue long su = 1; for (int i = 0; i < m; i++) { // Push the sum of all elements q.push(su); // Store front element of the queue long temp = q.front(); // Pop front from the queue q.pop(); // Set the value of su su = (2 * su) - temp; // print the ith N-bonacci number cout << temp << " "; }}// Driver's Codeint main(){ int N = 5, M = 15; bonacciseries(N, M); return 0;}// This code is contributed by Susobhan Akhuli |
Java
// Java program print first M terms of// N-bonacci series using queueimport java.util.*;public class bonacciSeries { // Function to find and print N-bonacci series static void bonacciseries(int n, long m) { Queue<Integer> q = new LinkedList<>(); // Assuming m > n. for (int i = 0; i < (n - 1); i++) { q.add(0); } // Nth Element in the queue q.add(1); // Initilize value of su as the sum of all the // elements, currently inside the queue long su = 1; for (int i = 0; i < m; i++) { // Push the sum of all elements q.add((int)su); // Store front element of the queue long temp = q.peek(); // Pop front from the queue q.remove(); // Set the value of su su = (2 * su) - temp; // print the ith N-bonacci number System.out.print(temp + " "); } } // Driver's Code public static void main(String[] args) { int N = 5, M = 15; bonacciseries(N, M); }}// This code is contributed by Susobhan Akhuli |
Output
0 0 0 0 1 1 2 4 8 16 31 61 120 236 464
Time Complexity: O(M), To iterate M times.
Auxiliary Space: O(N), because at every time queue has N elements.
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