Constrained Subsequence Sum

Problem Statement

The task is to find the maximum sum of a non-empty subsequence of an integer array nums, adhering to a specific condition based on the integer k. The condition is that for any two consecutive numbers in the chosen subsequence, their positions in the original array should not be more than k positions apart. The goal is to strategically select numbers from nums under this restriction to maximize the sum of the subsequence. Remember, a subsequence maintains the original order of elements but may skip over elements in the array.

Examples

Example 1

Input:

nums = [10,2,-10,5,20], k = 2

Output:

37
**Explanation:** The subsequence is [10, 2, 5, 20].

Example 2

Input:

nums = [-1,-2,-3], k = 1

Output:

-1
**Explanation:** The subsequence must be non-empty, so we choose the largest number.

Example 3

Input:

nums = [10,-2,-10,-5,20], k = 2

Output:

23
**Explanation:** The subsequence is [10, -2, -5, 20].

Constraints

• 1 <= k <= nums.length <= 105
• -104 <= nums[i] <= 104

Approach and Intuition

The problem essentially requires us to analyze segments or windows of the array and determine the best possible combinations of elements within these windows that can yield the highest sum.

1. Dynamic Programming as a Strategy: Given the constraints and the nature of the problem, dynamic programming (DP) is a promising approach to build solutions incrementally while ensuring adherence to the j - i <= k rule.
2. Utilizing a Sliding Window: A sliding window can help in managing the constraint j - i <= k, allowing us to continuously consider only those elements in the sequence that are allowed to be consecutively included based on k.
3. Decisions at Each Step: For every element in the array:
   • Determine whether including this element in the subsequence (while maintaining the distance constraint) would result in a higher sum than excluding it.
   • Update a DP table or similar structure that keeps track of the best possible sum up to the current element.
   • Manage maximum values within the window with the help of data structures like deques which can efficiently maintain maximum values over sliding window ranges.

Taking the given examples into account:

• Example 1 nums = [10,2,-10,5,20], k = 2: The allowable gap forces a selective addition primarily focusing on positive contributions while adhering to the index gap.
• Example 2 nums = [-1,-2,-3], k = 1: Emphasizes on choosing the least negative when all are negative and only single index gaps are allowed.
• Example 3 nums = [10,-2,-10,-5,20], k = 2: Here, strategic skipping of more severely negative numbers showcases the impact of choosing subsequences based on the permitted gaps.

The provided constraints suggest effectiveness and efficiency are necessary for solutions, given the extensive possible length of the array (k can be as large as 105). This makes the design of an efficacious algorithm using dynamic programming combined with sliding window techniques critical for managing both the sequence's subarray examination and subsequence summation efficiently.

Solutions

class Solution {
public:
    int maxSubsetSum(vector<int>& elements, int maximumDistance) {
        deque<int> indicesQueue;
        vector<int> computedMax(elements.size());
            
        for (int index = 0; index < elements.size(); index++) {
            if (!indicesQueue.empty() && index - indicesQueue.front() > maximumDistance) {
                indicesQueue.pop_front();
            }
                
            computedMax[index] = (!indicesQueue.empty() ? computedMax[indicesQueue.front()] : 0) + elements[index];
            while (!indicesQueue.empty() && computedMax[indicesQueue.back()] < computedMax[index]) {
                indicesQueue.pop_back();
            }
                
            if (computedMax[index] > 0) {
                indicesQueue.push_back(index);
            }
        }
            
        return *max_element(computedMax.begin(), computedMax.end());
    }
};