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 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)
# 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# 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 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)
?>
<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 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 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
// 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 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)
?>
<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 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 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
# 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# 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 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
Output0 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.
Explore
DSA Fundamentals
Data Structures
Algorithms
Advanced
Interview Preparation
Practice Problem