# Print the sequence of size N in which every term is sum of previous K terms

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:** First term of the series is **1** and if there are not enough previous terms then other terms are supposed to be **0**.**Examples:**

Input:N = 8, K = 3Output:1 1 2 4 7 13 24 44Explanation:

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 = 44Input:N = 10, K = 4Output: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++

`// C++ implementation to find the` `// series in which every term is` `// sum of previous K terms` `#include <iostream>` `using` `namespace` `std;` `// Function to generate the` `// series in the form of array` `void` `sumOfPrevK(` `int` `N, ` `int` `K)` `{` ` ` `int` `arr[N];` ` ` `arr[0] = 1;` ` ` `// Pick a starting point` ` ` `for` `(` `int` `i = 1; i < N; i++) {` ` ` `int` `j = i - 1, count = 0,` ` ` `sum = 0;` ` ` `// Find the sum of all` ` ` `// elements till count < K` ` ` `while` `(j >= 0 && count < K) {` ` ` `sum += arr[j];` ` ` `j--;` ` ` `count++;` ` ` `}` ` ` `// Find the value of` ` ` `// sum at i position` ` ` `arr[i] = sum;` ` ` `}` ` ` `for` `(` `int` `i = 0; i < N; i++) {` ` ` `cout << arr[i] << ` `" "` `;` ` ` `}` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `N = 10, K = 4;` ` ` `sumOfPrevK(N, K);` ` ` `return` `0;` `}` |

## Java

`// Java implementation to find the` `// series in which every term is` `// sum of previous K terms` `class` `Sum {` ` ` `// Function to generate the` ` ` `// series in the form of array` ` ` `void` `sumOfPrevK(` `int` `N, ` `int` `K)` ` ` `{` ` ` `int` `arr[] = ` `new` `int` `[N];` ` ` `arr[` `0` `] = ` `1` `;` ` ` `// Pick a starting point` ` ` `for` `(` `int` `i = ` `1` `; i < N; i++) {` ` ` `int` `j = i - ` `1` `, count = ` `0` `,` ` ` `sum = ` `0` `;` ` ` `// Find the sum of all` ` ` `// elements till count < K` ` ` `while` `(j >= ` `0` `&& count < K) {` ` ` `sum += arr[j];` ` ` `j--;` ` ` `count++;` ` ` `}` ` ` `// Find the value of` ` ` `// sum at i position` ` ` `arr[i] = sum;` ` ` `}` ` ` `for` `(` `int` `i = ` `0` `; i < N; i++) {` ` ` `System.out.print(arr[i] + ` `" "` `);` ` ` `}` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `main(String args[])` ` ` `{` ` ` `Sum s = ` `new` `Sum();` ` ` `int` `N = ` `10` `, K = ` `4` `;` ` ` `s.sumOfPrevK(N, K);` ` ` `}` `}` |

## Python3

`# Python3 implementation to find the` `# series in which every term is` `# sum of previous K terms` `# Function to generate the` `# series in the form of array` `def` `sumOfPrevK(N, K):` ` ` `arr ` `=` `[` `0` `for` `i ` `in` `range` `(N)]` ` ` `arr[` `0` `] ` `=` `1` ` ` `# Pick a starting point` ` ` `for` `i ` `in` `range` `(` `1` `,N):` ` ` `j ` `=` `i ` `-` `1` ` ` `count ` `=` `0` ` ` `sum` `=` `0` ` ` ` ` `# Find the sum of all` ` ` `# elements till count < K` ` ` `while` `(j >` `=` `0` `and` `count < K):` ` ` `sum` `=` `sum` `+` `arr[j]` ` ` `j ` `=` `j ` `-` `1` ` ` `count ` `=` `count ` `+` `1` ` ` `# Find the value of` ` ` `# sum at i position` ` ` `arr[i] ` `=` `sum` ` ` `for` `i ` `in` `range` `(` `0` `, N):` ` ` `print` `(arr[i])` `# Driver Code` `N ` `=` `10` `K ` `=` `4` `sumOfPrevK(N, K)` `# This code is contributed by Sanjit_Prasad` |

## C#

`// C# implementation to find the` `// series in which every term is` `// sum of previous K terms` `using` `System;` `class` `Sum {` ` ` `// Function to generate the` ` ` `// series in the form of array` ` ` `void` `sumOfPrevK(` `int` `N, ` `int` `K)` ` ` `{` ` ` `int` `[]arr = ` `new` `int` `[N];` ` ` `arr[0] = 1;` ` ` ` ` `// Pick a starting point` ` ` `for` `(` `int` `i = 1; i < N; i++) {` ` ` `int` `j = i - 1, count = 0,` ` ` `sum = 0;` ` ` `// Find the sum of all` ` ` `// elements till count < K` ` ` `while` `(j >= 0 && count < K) {` ` ` `sum += arr[j];` ` ` `j--;` ` ` `count++;` ` ` `}` ` ` `// Find the value of` ` ` `// sum at i position` ` ` `arr[i] = sum;` ` ` `}` ` ` `for` `(` `int` `i = 0; i < N; i++) {` ` ` `Console.Write(arr[i] + ` `" "` `);` ` ` `}` ` ` `}` ` ` ` ` `// Driver Code` ` ` `public` `static` `void` `Main(String []args)` ` ` `{` ` ` `Sum s = ` `new` `Sum();` ` ` `int` `N = 10, K = 4;` ` ` `s.sumOfPrevK(N, K);` ` ` `}` `}` `// This code is contributed by 29AjayKumar` |

## Javascript

`<script>` `// JavaScript implementation to find the` `// series in which every term is` `// sum of previous K terms` `// Function to generate the` `// series in the form of array` `function` `sumOfPrevK(N, K)` `{` ` ` `let arr = ` `new` `Array(N);` ` ` `arr[0] = 1;` ` ` `// Pick a starting point` ` ` `for` `(let i = 1; i < N; i++) {` ` ` `let j = i - 1, count = 0, sum = 0;` ` ` `// Find the sum of all` ` ` `// elements till count < K` ` ` `while` `(j >= 0 && count < K) {` ` ` `sum += arr[j];` ` ` `j--;` ` ` `count++;` ` ` `}` ` ` `// Find the value of` ` ` `// sum at i position` ` ` `arr[i] = sum;` ` ` `}` ` ` `for` `(let i = 0; i < N; i++) {` ` ` `document.write(arr[i] + ` `" "` `);` ` ` `}` `}` `// Driver Code` `let N = 10, K = 4;` `sumOfPrevK(N, K);` `// This code is contributed by _saurabh_jaiswal` `</script>` |

**Output:**

1 1 2 4 8 15 29 56 108 208

**Performance analysis:**

**Time Complexity:**O(N * K)**Space Complexity:**O(N)

**Efficient Approach:** The idea is to store the current sum in a variable and in every step subtract the last **K ^{th}** term 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++

`// C++ implementation to find the` `// series in which every term is` `// sum of previous K terms` `#include <iostream>` `using` `namespace` `std;` `// Function to generate the` `// series in the form of array` `void` `sumOfPrevK(` `int` `N, ` `int` `K)` `{` ` ` `int` `arr[N], prevsum = 0;` ` ` `arr[0] = 1;` ` ` `// Pick a starting point` ` ` `for` `(` `int` `i = 0; i < N - 1; i++) {` ` ` `// Computing the previous sum` ` ` `if` `(i < K) {` ` ` `arr[i + 1] = arr[i] + prevsum;` ` ` `prevsum = arr[i + 1];` ` ` `}` ` ` `else` `{` ` ` `arr[i + 1] = arr[i] + prevsum` ` ` `- arr[i + 1 - K];` ` ` `prevsum = arr[i + 1];` ` ` `}` ` ` `}` ` ` `// Loop to print the series` ` ` `for` `(` `int` `i = 0; i < N; i++) {` ` ` `cout << arr[i] << ` `" "` `;` ` ` `}` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `N = 8, K = 3;` ` ` `sumOfPrevK(N, K);` ` ` `return` `0;` `}` |

## Java

`// Java implementation to find the` `// series in which every term is` `// sum of previous K terms` `class` `Sum {` ` ` `// Function to generate the` ` ` `// series in the form of array` ` ` `void` `sumOfPrevK(` `int` `N, ` `int` `K)` ` ` `{` ` ` `int` `arr[] = ` `new` `int` `[N];` ` ` `int` `prevsum = ` `0` `;` ` ` `arr[` `0` `] = ` `1` `;` ` ` `// Pick a starting point` ` ` `for` `(` `int` `i = ` `0` `; i < N - ` `1` `; i++) {` ` ` `// Computing the previous sum` ` ` `if` `(i < K) {` ` ` `arr[i + ` `1` `] = arr[i] + prevsum;` ` ` `prevsum = arr[i + ` `1` `];` ` ` `}` ` ` `else` `{` ` ` `arr[i + ` `1` `] = arr[i] + prevsum` ` ` `- arr[i + ` `1` `- K];` ` ` `prevsum = arr[i + ` `1` `];` ` ` `}` ` ` `}` ` ` `// Loop to print the series` ` ` `for` `(` `int` `i = ` `0` `; i < N; i++) {` ` ` `System.out.print(arr[i] + ` `" "` `);` ` ` `}` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `main(String args[])` ` ` `{` ` ` `Sum s = ` `new` `Sum();` ` ` `int` `N = ` `8` `, K = ` `3` `;` ` ` `s.sumOfPrevK(N, K);` ` ` `}` `}` |

## Python3

`# Python3 implementation to find the` `# series in which every term is` `# sum of previous K terms` `# Function to generate the` `# series in the form of array` `def` `sumOfPrevK(N, K):` ` ` `arr ` `=` `[` `0` `]` `*` `N;` ` ` `prevsum ` `=` `0` `;` ` ` `arr[` `0` `] ` `=` `1` `;` ` ` `# Pick a starting point` ` ` `for` `i ` `in` `range` `(N` `-` `1` `):` ` ` `# Computing the previous sum` ` ` `if` `(i < K):` ` ` `arr[i ` `+` `1` `] ` `=` `arr[i] ` `+` `prevsum;` ` ` `prevsum ` `=` `arr[i ` `+` `1` `];` ` ` `else` `:` ` ` `arr[i ` `+` `1` `] ` `=` `arr[i] ` `+` `prevsum ` `-` `arr[i ` `+` `1` `-` `K];` ` ` `prevsum ` `=` `arr[i ` `+` `1` `];` ` ` `# Loop to print the series` ` ` `for` `i ` `in` `range` `(N):` ` ` `print` `(arr[i], end` `=` `" "` `);` ` ` `# Driver code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `N ` `=` `8` `;` ` ` `K ` `=` `3` `;` ` ` `sumOfPrevK(N, K);` ` ` `# This code is contributed by 29AjayKumar` |

## C#

`// C# implementation to find the` `// series in which every term is` `// sum of previous K terms` `using` `System;` `public` `class` `Sum {` ` ` `// Function to generate the` ` ` `// series in the form of array` ` ` `void` `sumOfPrevK(` `int` `N, ` `int` `K)` ` ` `{` ` ` `int` `[]arr = ` `new` `int` `[N];` ` ` `int` `prevsum = 0;` ` ` `arr[0] = 1;` ` ` ` ` `// Pick a starting point` ` ` `for` `(` `int` `i = 0; i < N - 1; i++) {` ` ` ` ` `// Computing the previous sum` ` ` `if` `(i < K) {` ` ` `arr[i + 1] = arr[i] + prevsum;` ` ` `prevsum = arr[i + 1];` ` ` `}` ` ` `else` `{` ` ` `arr[i + 1] = arr[i] + prevsum` ` ` `- arr[i + 1 - K];` ` ` `prevsum = arr[i + 1];` ` ` `}` ` ` `}` ` ` ` ` `// Loop to print the series` ` ` `for` `(` `int` `i = 0; i < N; i++) {` ` ` `Console.Write(arr[i] + ` `" "` `);` ` ` `}` ` ` `}` ` ` ` ` `// Driver code` ` ` `public` `static` `void` `Main(String []args)` ` ` `{` ` ` `Sum s = ` `new` `Sum();` ` ` `int` `N = 8, K = 3;` ` ` `s.sumOfPrevK(N, K);` ` ` `}` `}` `// This code is contributed by 29AjayKumar` |

**Complexity analysis:**

**Time Complexity:**O(N)**Space Complexity:**O(N)

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