Constrained Subsequence Sum

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());
    }
};

The provided C++ code implements a solution to the "Constrained Subsequence Sum" problem. Here's a summary of how this implementation works:

• Define a function maxSubsetSum which takes a vector of integers elements and an integer maximumDistance as parameters.
• Utilize a deque to efficiently manage indices of the elements vector that are within the specified maximum distance and can potentially form the maximum sum.
• Create a vector computedMax where each element at index i stores the maximum possible sum of a subsequence ending at i that abides by the distance constraint.
• Loop through each element in elements:
  • Remove indices from the front of the deque that are not within the allowed distance of maximumDistance.
  • Calculate computedMax for the current index as the sum of the elements[index] and the maximum sum ending within the distance (or 0 if the deque is empty).
  • Use the deque to maintain a list of indices with potential maximal sums, ensuring it stores only the indices leading to the highest values in a decreasing order.
  • Place the current index in indicesQueue if the computed maximum sum at this index is positive.
• After processing all elements, determine the global maximum sum by finding the maximum value in computedMax.

The algorithm effectively reduces unnecessary computations by discarding any index that either falls outside the distance constraint or does not contribute to a potential maximum due to a lower computedMax value compared to newer indices. This results in a solution that balances between time efficiency and computing the requisite maximum sum respecting the constraints posed by maximumDistance. This type of problem is primarily approached using dynamic programming combined with a deque to achieve optimal time complexity.

class Solution {
    public int maximumSubsetSum(int[] arr, int limit) {
        Deque<Integer> deque = new ArrayDeque<>();
        int[] result = new int[arr.length];

        for (int index = 0; index < arr.length; index++) {
            if (!deque.isEmpty() && index - deque.getFirst() > limit) {
                deque.pollFirst();
            }

            result[index] = (deque.isEmpty() ? 0 : result[deque.getFirst()]) + arr[index];
            
            while (!deque.isEmpty() && result[deque.getLast()] < result[index]) {
                deque.removeLast();
            }

            if (result[index] > 0) {
                deque.addLast(index);
            }
        }

        int maxSum = Integer.MIN_VALUE;
        for (int res : result) {
            maxSum = Math.max(maxSum, res);
        }

        return maxSum;
    }
}

To solve the Constrained Subsequence Sum problem in Java, implement the provided algorithm that efficiently computes the maximum sum of a subsequence where the distance between indices is constrained by a specified limit. Here's a detailed breakdown of how the solution works:

• Use a Deque (double-ended queue) to efficiently maintain potential maximum values within the specified limit.
• Create an array result to store the maximum sum that ends at each index of the input array arr.
• Loop through each element of arr. For each element at index:
  • Remove indices from the front of the deque if they are out of the specified limit.
  • Calculate the maximum sum that can end at index by adding the current element arr[index] to the maximum sum ending at the index stored at the front of the deque.
  • Remove elements from the back of the deque as long as the sum at index is greater than the sums at those indexes, ensuring that deque always has the largest sums at the front.
  • If the result at the current index is positive, add that index to the deque.
• After processing all elements, loop through the result array to find the maximum sum.

Use this approach to ensure an optimal solution with time complexity primarily dominated by the linear pass over arr, thus offering both efficiency and simplicity. This algorithm is suitable for large inputs constrained by both value and index-wise distance.

class Solution:
    def maxSubsetSum(self, numbers: List[int], limit: int) -> int:
        deq = deque()
        results = [0] * len(numbers)

        for index in range(len(numbers)):
            if deq and index - deq[0] > limit:
                deq.popleft()

            results[index] = (results[deq[0]] if deq else 0) + numbers[index]
            while deq and results[deq[-1]] < results[index]:
                deq.pop()

            if results[index] > 0:
                deq.append(index)

        return max(results)

This solution implements an approach to solve the "Constrained Subsequence Sum" problem using Python. The algorithm optimally finds the maximum sum of a subsequence in an array of integers where the length of the subsequence is constrained by a given limit. Here's how the solution works:

• A deque deq and a list results are initialized. deq helps maintain indices of array elements that contribute to the maximum sum, ensuring quick updates and constraint checks. results stores the maximum sums obtainable at each position within the constraints.
• Loop through each number in the input list numbers. For each number at position index:
  1. Check if the first element in the deque is outside the constrained range (older than limit). If so, remove it from the deque.
  2. Calculate the current maximum possible value for results[index] using the highest value from results indexed at the start of the deque added to the current number.
  3. Maintain the deque in descending order of values stored in results. Remove elements from the end of the deque if they are less than the new result calculated.
  4. Only indices whose results are greater than 0 are considered for future sums and thus appended to the deque.
• After iterating through all elements, return the maximum value from results.

This implementation efficiently handles the limits on subsequence length using a monotonic deque to both maintain the possible maximum sums and discard old values that are no longer within the allowed range, ensuring a time complexity more favorable than using nested loops.