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Given two integers N and K, the task is to generate a series of N terms in which every term is sum of the previous K terms.
Note: The first term of the series is
1. if there are not enough previous terms, then other terms are supposed to be 0.
Examples:
Input: N = 8, K = 3
Output: 1 1 2 4 7 13 24 44
Explanation:
Series is generated as follows:
a[0] = 1
a[1] = 1 + 0 + 0 = 1
a[2] = 1 + 1 + 0 = 2
a[3] = 2 + 1 + 1 = 4
a[4] = 4 + 2 + 1 = 7
a[5] = 7 + 4
+ 2 = 13
a[6] = 13 + 7 + 4 = 24
a[7] = 24 + 13 + 7 = 44
Input: N = 10, K = 4
Output: 1 1 2 4 8 15 29 56 108 208
Naive Approach: The idea is to run two loops to generate N terms of series. Below is the illustration of the steps:
- Traverse the first loop from 0 to N – 1, to generate every term of the series.
- Run a loop from
max(0, i – K) to i to compute the sum of the previous K terms.
- Update the sum to the current index of series as the current term.
Below is the implementation of the above approach:
C++
#include
using namespace std;
void sumOfPrevK(int N, int K)
{
int arr[N];
arr[0] = 1;
for (int i = 1; i < N; i++) {
int
j = i - 1, count = 0,
sum = 0;
while (j >= 0 && count < K) {
sum += arr[j];
j--;
count++;
}
arr[i] = sum;
}
for (int i = 0; i < N; i++) {
cout << arr[i] << " ";
}
}
int main()
{
int N = 10, K = 4;
sumOfPrevK(N, K);
return 0;
}
Java
class
Sum {
void sumOfPrevK(int N, int K)
{
int arr[] = new int[N];
arr[0] = 1;
for (int i = 1; i < N; i++) {
int j = i - 1, count = 0,
sum = 0;
while (j >= 0 && count < K) {
sum += arr[j];
j--;
count++;
}
arr[i] = sum;
}
for (int i = 0; i < N; i++) {
System.out.print(arr[i] + " ");
}
}
public static void main(String args[])
{
Sum s = new Sum();
int N = 10, K = 4;
s.sumOfPrevK(N, K);
}
}
Python3
def sumOfPrevK(N, K):
arr = [0 for i in range(N)]
arr[0] = 1
for i in range(1,N):
j = i - 1
count = 0
sum = 0
while (j >= 0 and count < K):
sum = sum + arr[j]
j =
j - 1
count = count + 1
arr[i] = sum
for i in range(0, N):
print(arr[i])
N = 10
K = 4
sumOfPrevK(N, K)
C#
using System;
class Sum {
void sumOfPrevK(int N, int K)
{
int []arr = new
int[N];
arr[0] = 1;
for (int i = 1; i < N; i++) {
int j = i - 1, count = 0,
sum = 0;
while (j >= 0 && count < K) {
sum += arr[j];
j--;
count++;
}
arr[i] = sum;
}
for (int i = 0; i < N; i++) {
Console.Write(arr[i] + " ");
}
}
public static void
Main(String []args)
{
Sum s = new Sum();
int N = 10, K = 4;
s.sumOfPrevK(N, K);
}
}
Javascript
Output:
1 1 2 4 8 15 29 56 108 208
Performance analysis:
- Time Complexity: O(N * K)
- Space Complexity: O(N)
Improved Approach: The idea is to store the current sum in a variable and
subtract the last Kth term in every step and add the last term into the pre-sum to compute every term of the series.
Below is the implementation of the above approach:
C++
#include
#include
using namespace std;
void sumOfPrevK(int N, int K)
{
vector<int> arr(N);
arr[0] = 1;
int start = 0, end = 1, sum = 1;
while (end < N) {
if
(end - start > K) {
sum -= arr[start];
++start;
}
arr[end] = sum;
sum += arr[end];
++end;
}
for (int i = 0; i < N; ++i) {
cout << arr[i] << " ";
}
cout << endl;
}
int main()
{
int N = 10, K = 4;
sumOfPrevK(N, K);
return 0;
}
Java
import java.util.*;
class GFG {
static void sumOfPrevK(int N, int K)
{
ArrayList arr = new ArrayList();
for (int i = 0; i < N; i++)
arr.add(0);
arr.set(0, 1);
int start = 0, end = 1, sum = 1;
while (end < N) {
if (end - start > K) {
sum -= arr.get(start);
++start;
}
arr.set(end, sum);
sum += arr.get(end);
++end;
}
for (int i = 0; i < N; ++i) {
System.out.print(arr.get(i) + " ");
}
System.out.print("\n");
}
public static void main(String[] args)
{
int N = 10, K = 4;
sumOfPrevK(N, K);
}
}
Python3
def sumsOfPrevK(N, K):
arr = [0 for _ in range(N)]
arr[0] = 1
start = 0
end = 1
sums = 1
while (end < N):
if (end - start > K):
sums -= arr[start]
start += 1
arr[end] = sums
sums += arr[end]
end += 1
print(*arr)
N = 10
K = 4
sumsOfPrevK(N, K)
C#
using System;
using System.Collections.Generic;
class GFG
{
static void sumOfPrevK(int N, int K)
{
List<int> arr = new
List<int>() ;
for (int i = 0; i < N; i++)
arr.Add(0);
arr[0] = 1;
int start = 0, end = 1, sum = 1;
while (end < N) {
if (end - start > K) {
sum -= arr[start];
++start;
}
arr[end] = sum;
sum += arr[end];
++end;
}
for (int i = 0; i < N; ++i) {
Console.Write(arr[i] + " ");
}
Console.Write("\n");
}
public static void Main(string[] args)
{
int N = 10, K = 4;
sumOfPrevK(N, K);
}
}
Javascript
function sumOfPrevK(N, K)
{
let arr = new Array(N);
arr[0] = 1;
let start = 0, end = 1, sum = 1;
while (end < N) {
if (end - start > K) {
sum -= arr[start];
++start;
}
arr[end] = sum;
sum += arr[end];
++end;
}
console.log(arr.join(" "))
}
let N = 10, K = 4;
sumOfPrevK(N, K);
Complexity analysis:
- Time Complexity: O(N)
- Space Complexity: O(N)
Another Approach: Sliding Window
The idea is to use a fixed-size sliding window of size K to get the sum of previous K
elements.
- Initialize the answer array with arr[0] = 1 and a variable sum = 1, representing the sum of the last K elements. Initially, the previous sum = 1. (If there are less than K elements in the last window, then the sum will store the sum of all the elements in the window)
- The sliding window is initialized by making start = 0 and end = 1.
- If
the window size exceeds K, then decrement arr[start] from sum and increment start toupdate the window size and sum.
- Make arr[end] = sum, which will be the sum of the previous K elements, and increment sum by arr[end] for the next iteration. Increment end by 1.
Output
1 1 2 4 8 15 29 56 108 208
Complexity
analysis:
- Time Complexity: O(N)
- Auxiliary Space: O(N)
JavaScript does not have any print object or print methods. You cannot access output devices from JavaScript. The only exception is that you can call the window.print() method in the browser to print the content of the current window.
const num1 = parseInt(prompt('Enter the first number ')); const num2 = parseInt(prompt('Enter the second
number ')); Then, the sum of the numbers is computed. const sum = num1 + num2; Finally, the sum is displayed.
Javascript Number() Function object: This parameter holds the objects that will be converted any type of javascript variable to number type. Return Values: Number() function returns the number format for any type of javascript variable. Example 2: Not a number is returned by the compiler.
