Given an array and a number k, find a subsequence such that

- Sum of elements in subsequence is maximum
- Indices of elements of subsequence differ atleast by k
- If we select element at index i such that i + k + 1 >= N, then we cannot select any other element as part of the subsequence. Hence we need to decide whether to select this element or one of the elements after it.
- If we select element at index i such that i + k + 1 < N, then the next element we can select is at i + k + 1 index. Thus we need to decide whether to select these two elements, or move on to the next adjacent element.

Examples

Input : arr[] = {4, 5, 8, 7, 5, 4, 3, 4, 6, 5} k = 2 Output: 19 Explanation: The highest value is obtained if you pick indices 1, 4, 7, 10 giving 4 + 7 + 3 + 5 = 19 Input: arr[] = {50, 70, 40, 50, 90, 70, 60, 40, 70, 50} k = 2 Output: 230 Explanation: There are 10 elements and k = 2. If you select 2, 5, and 9 you get a total value of 230, which is the maximum possible.

A **simple solution **is to consider all subsequences one by one. In every subsequence, check for distance condition and return the maximum sum subsequence.

An **efficient solution** is to use dynamic programming.

There are two cases:

These two cases can be written as:

Let MS[i] denotes the maximum sum of subsequence from i to N. Base Case: MS[N-1] = arr[N-1] If i + 1 + k >= N MS[i] = max(arr[i], MS[i+1]), Else MS[i] = max(arr[i] + MS[i+k+1], MS[i+1]) Evidently, the solution to the problem is to find MS[0].

Below is the implementation:

## C++

// CPP program to find maximum sum subsequence // such that elements are at least k distance // away. #include <bits/stdc++.h> using namespace std; int maxSum(int arr[], int N, int k) { // MS[i] is going to store maximum sum // subsequence in subarray from arr[i] // to arr[n-1] int MS[N]; // We fill MS from right to left. MS[N - 1] = arr[N - 1]; for (int i = N - 2; i >= 0; i--) { if (i + k + 1 >= N) MS[i] = max(arr[i], MS[i + 1]); else MS[i] = max(arr[i] + MS[i + k + 1], MS[i + 1]); } return MS[0]; } // Driver code int main() { int N = 10, k = 2; int arr[] = { 50, 70, 40, 50, 90, 70, 60, 40, 70, 50 }; cout << maxSum(arr, N, k); return 0; }

## Java

// Java program to find maximum sum subsequence // such that elements are at least k distance // away. import java.io.*; class GFG { static int maxSum(int arr[], int N, int k) { // MS[i] is going to store maximum sum // subsequence in subarray from arr[i] // to arr[n-1] int MS[] = new int[N]; // We fill MS from right to left. MS[N - 1] = arr[N - 1]; for (int i = N - 2; i >= 0; i--) { if (i + k + 1 >= N) MS[i] = Math.max(arr[i], MS[i + 1]); else MS[i] = Math.max(arr[i] + MS[i + k + 1], MS[i + 1]); } return MS[0]; } // Driver code public static void main(String[] args) { int N = 10, k = 2; int arr[] = { 50, 70, 40, 50, 90, 70, 60, 40, 70, 50 }; System.out.println(maxSum(arr, N, k)); } } // This code is contributed by Prerna Saini

Output:

230

Time Complexity : O(n)

Auxiliary Space : O(n)

This article is contributed by **Sayan Mahapatra**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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