# Maximum sum subsequence with at-least k distant elements

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.
- Maximum length subsequence such that adjacent elements in the subsequence have a common factor
- Maximum length subsequence with difference between adjacent elements as either 0 or 1
- Minimal product subsequence where adjacent elements are separated by a maximum distance of K
- Maximum average of a subarray of size of atleast X and atmost Y
- Subsequence with maximum odd sum
- Maximum even sum subsequence
- Subsequence of size k with maximum possible GCD
- Maximum Sum Decreasing Subsequence
- Maximum Sum Increasing Subsequence | DP-14
- Maximum subsequence sum such that no three are consecutive
- Maximum sum alternating subsequence
- Printing Maximum Sum Increasing Subsequence
- Maximum product of subsequence of size k
- Maximum product of an increasing subsequence
- Maximum product of an increasing subsequence of size 3

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 = N-2 to 0. 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; ` `}` |

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## 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 ` |

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## Python3

` ` `# Python3 program to find maximum ` `# sum subsequence such that elements ` `# are at least k distance away. ` ` ` `def` `maxSum(arr, N, k): ` ` ` ` ` `# MS[i] is going to store maximum sum ` ` ` `# subsequence in subarray from arr[i] ` ` ` `# to arr[n-1] ` ` ` `MS ` `=` `[` `0` `for` `i ` `in` `range` `(N)] ` ` ` ` ` `# We fill MS from right to left. ` ` ` `MS[N ` `-` `1` `] ` `=` `arr[N ` `-` `1` `] ` ` ` `for` `i ` `in` `range` `(N ` `-` `2` `, ` `-` `1` `, ` `-` `1` `): ` ` ` `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 ` `N ` `=` `10` `; k ` `=` `2` `arr ` `=` `[ ` `50` `, ` `70` `, ` `40` `, ` `50` `, ` `90` `, ` `70` `, ` `60` `, ` `40` `, ` `70` `, ` `50` `] ` `print` `(maxSum(arr, N, k)) ` ` ` `# This code is contributed by Anant Agarwal. ` |

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## C#

`// C# program to find maximum sum ` `// subsequence such that elements ` `// are at least k distance away. ` `using` `System; ` ` ` `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() ` ` ` `{ ` ` ` `int` `N = 10, k = 2; ` ` ` `int` `[]arr = { 50, 70, 40, 50, 90, 70, 60, ` ` ` `40, 70, 50 }; ` ` ` `Console.WriteLine(maxSum(arr, N, k)); ` ` ` `} ` `} ` ` ` `// This code is contributed by Anant Agarwal. ` |

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## PHP

`<?php ` `// PHP program to find ` `// maximum sum subsequence ` `// such that elements are ` `// at least k distance ` `// away. ` ` ` `function` `maxSum(` `$arr` `, ` `$N` `, ` `$k` `) ` `{ ` ` ` ` ` `// MS[i] is going to ` ` ` `// store maximum sum ` ` ` `// subsequence in ` ` ` `// subarray from arr[i] ` ` ` `// to arr[n-1] ` ` ` ` ` `// We fill MS from ` ` ` `// right to left. ` ` ` `$MS` `[` `$N` `- 1] = ` `$arr` `[` `$N` `- 1]; ` ` ` `for` `(` `$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 ` `$N` `= 10; ` `$k` `= 2; ` `$arr` `= ` `array` `(50, 70, 40, 50, 90, ` ` ` `70, 60, 40, 70, 50); ` `echo` `(maxSum(` `$arr` `, ` `$N` `, ` `$k` `)); ` ` ` `// This code is contributed by Ajit. ` `?> ` |

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Output:

230

Time Complexity : O(n)

Auxiliary Space : O(n)

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