Maximum sum subsequence with at-least k distant elements

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

    1. Sum of elements in subsequence is maximum
    2. Indices of elements of subsequence differ atleast by k

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.

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:

        1. 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.
        2. 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.

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:

// 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;
}

 

      1. Output:
230

Time Complexity : O(n)
Auxiliary Space : O(n)

Disclaimer: This does not belong to TechCodeBit, its an article taken from the below
source and credits.
source and credits: http://www.geeksforgeeks.org
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rakesh

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