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 series
void 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 Code
int 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 series


def 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 Code
N = 5
M = 15
bonacciseries(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 series
function 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 series
void 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 Code
int 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 series


def 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 Code
if __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 series
function 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:

1. First we push N element inside the queue whose values will be ‘0‘ for 1 to N-1 elements and ‘1‘ for Nth element.
2. 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.
3. 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 series
void 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);

    // Initialize 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 Code
int 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 queue
import 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);

        // Initialize 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

Python3

# Python program print first M terms of
# N-bonacci series using queue
from queue import Queue

# Function to find and print N-bonacci series
def bonacciseries(n, m):
    q = Queue()
    # Assuming m > n.
    for i in range(n - 1):
        q.put(0)

    # Nth Element in the queue
    q.put(1)

    # Initialize value of su as the sum of all the elements,
    # currently inside the queue
    su = 1

    for i in range(m):

        # Push the sum of all elements
        q.put(su)

        # Store front element of the queue
        temp = q.queue[0]

        # Pop front from the queue
        q.get()

        # Set the value of su
        su = (2 * su) - temp

        # print the ith N-bonacci number
        print(temp, end=" ")

# Driver's Code
if __name__ == "__main__":
    N, M = 5, 15
    bonacciseries(N, M)

# This code is contributed by Susobhan Akhuli

C#

// C# program print first M terms of
// N-bonacci series using queue
using System;
using System.Collections.Generic;

class GFG {
  // Function to find and print N-bonacci series
  static void bonacciseries(int n, long m)
  {
    Queue<int> q = new Queue<int>();
    // Assuming m > n.
    for (int i = 0; i < (n - 1); i++) {
      q.Enqueue(0);
    }

    // Nth Element in the queue
    q.Enqueue(1);

    // Initialize 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.Enqueue((int)su);

      // Store front element of the queue
      int temp = q.Peek();

      // Pop front from the queue
      q.Dequeue();

      // Set the value of su
      su = (2 * su) - temp;

      // print the ith N-bonacci number
      Console.Write(temp + " ");
    }
  }

  // Driver's Code
  static void Main()
  {
    int N = 5, M = 15;
    bonacciseries(N, M);
  }
}

// This code is contributed by Susobhan Akhuli

Javascript

// JavaScript program print first M terms of
// N-bonacci series using queue
function bonacciseries(n, m) {
    var q = [];

    //const q = new Queue();
    // Assuming m > n.
    for (let i = 0; i < (n - 1); i++) {
        q.push(0);
    }

    // Nth Element in the queue
    q.push(1);

    // Initialize value of su as the sum of all the
    // elements, currently inside the queue
    let su = 1;

    for (let i = 0; i < m; i++) {
        // Push the sum of all elements
        q.push(su);

        // Store front element and pop it from the queue
        let temp = q.shift();

        // Set the value of su
        su = (2 * su) - temp;

        // print the ith N-bonacci number
        console.log(temp + " ");
    }
}

// Driver's Code
let 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.