Problem Description

Given an integer array nums and an integer k, you are tasked with finding the maximum sum of a non-empty subsequence within the array. A subsequence here is defined as a sequence that can be derived from the array by deleting some elements without changing the order of the remaining elements. However, there's an additional constraint with regard to the subsequence. For any two consecutive integers in the subsequence, say nums[i] and nums[j] where i < j, the difference between the indices j - i must be less than or equal to k.

In simplified terms, you need to choose a subsequence such that any two adjacent numbers in this subsequence are not more than k positions apart in the original array, and the sum of this subsequence is as large as possible.

Intuition

To arrive at the solution, consider two important aspects: dynamic programming (DP) for keeping track of the subsequences and a 'sliding window' to enforce the constraint of the elements being within a distance k of each other.

With dynamic programming, create an array dp where each dp[i] stores the maximum sum of the subsequence up to the i-th element by following the constraint. To maintain the k distance constraint, use a deque (double-ended queue) that acts like a sliding window, carrying indices of elements that are potential candidates for the maximum sum. At each step:

• Remove indices from the front of the deque which are out of the current window of size k.
• Calculate dp[i] by adding the current element nums[i] to the max sum of the window (which is dp[q[0]], q being the deque). If there are no elements in q, just take the current element nums[i].
• Since we want to maintain a decreasing order of dp values in the deque to quickly access the maximum sum, remove elements from the end of the deque which have a dp value less than or equal to the current dp[i].
• Add the current index i to the deque.
• Update the answer with the maximum dp[i] seen so far.

This approach ensures that the deque always contains indices whose dp values form a decreasing sequence, and thus the front of the deque gives us the maximum sum for the current window, while satisfying the distance constraint.

Learn more about Queue, Dynamic Programming, Sliding Window, Monotonic Queue and Heap (Priority Queue) patterns.

Solution Approach

The implementation uses dynamic programming in combination with a deque to efficiently solve the problem by remembering previous results and ensuring that the constraint is maintained.

The initial setup involves creating a DP table as a list dp of size n (the length of nums) and initializing an answer variable ans to negative infinity to keep track of the maximum subsequence sum found so far.

Here is a step-by-step walkthrough of the implementation:

1. Iterate over the array nums using the index i and the value v. This loop will determine the dp[i] value for each element in nums.

2. If the deque q has elements and the oldest element's index in the deque (q[0]) is out of the window (i.e., i - q[0] > k), pop it from the front of the deque.

3. Set dp[i] to the maximum of 0 and the sum of the value v added to dp[q[0]] (the maximum sum within the window). If q is empty, just add v to 0. This step ensures that we do not consider subsequences with negative sums since they would decrease the overall maximum sum we are trying to compute.

4. While the deque has elements and the last element in the deque has a dp value less than or equal to dp[i] (since dp[i] now holds the maximum sum up to i), pop elements from the end of the deque. This maintains the invariant that the deque holds indices in decreasing order of their dp values.

5. Append the current index i to the deque.

6. Update ans with the maximum value between ans and dp[i] to update the global maximum sum each time dp[i] is calculated.

7. Once the iteration over nums is complete, ans will contain the maximum sum of a subsequence satisfying the given constraint, so return ans.

This algorithm makes use of the following patterns and data structures:

• Dynamic Programming: By storing and reusing solutions to subproblems (dp[i]), we solve larger problems efficiently.
• Monotonic Queue (Deque): A deque allows us to efficiently track the "window" of elements that are within k distance from the current element while simultaneously maintaining a monotonically decreasing order of dp values.
• Sliding Window Technique: The concept of having a window moving through the data while maintaining certain conditions (here, a maximum sum within a distance k) is a classic example of this technique.

Example Walkthrough

Let's assume we have an integer array nums = [10, 2, -10, 5, 20] and k = 2, and we want to find the maximum sum of a non-empty subsequence such that consecutive elements in the subsequence are not more than k positions apart in the original array.

We initialize our dp and ans:

• dp = [0, 0, 0, 0, 0] with the same length as nums
• ans = -∞ (minimum possible value)

We start by iterating over nums:

1. For i = 0, v = 10. The deque q is empty, so dp[0] = 10 and q = [0]. Now ans = 10.
2. For i = 1, v = 2. Since q[0] is 0 and i - q[0] = 1 which is within the range k, we calculate dp[1] = max(dp[q[0]] + v, v) = max(10 + 2, 2) = 12. Now q = [0, 1] after cleaning outnumbers by dp value (no changes in this step), and ans = 12.
3. For i = 2, v = -10. Since i - q[0] = 2 which is within k, we calculate dp[2] = max(dp[q[0]] + v, v) = max(12 - 10, -10) but here both options are negative, so dp[2] = 0. We then remove from q since dp[1] > dp[2]. Now q = [1] and ans = 12.
4. For i = 3, v = 5. Since i - q[0] = 2 which is equal to k, we can use q[0]. Thus, dp[3] = max(dp[q[0]] + v, v) = max(12 + 5, 5) = 17 and q = [1, 3]. We update ans = 17 now.
5. For i = 4, v = 20. Here, we first remove q[0] because i - q[0] = 3 which is greater than k, leaving q = [3]. We then calculate dp[4] = dp[q[0]] + v = 17 + 20 = 37, so q = [4] after removing q[0] because dp[3] < dp[4]. ans is updated to 37.

After iterating through the loop, we've looked at all possible subsequences that obey the k limit rule. The maximum subsequence sum is stored in ans which is 37.

To summarize this walk-through:

• We used DP to remember the maximum subsequence sums for subarrays.
• We maintained a deque q to ensure elements are within k distance in the subsequence.
• We updated ans at every step with the maximum value of dp[i].
• At the end of the loop, ans gave us the maximum sum possible under the given constraints.

Solution Implementation

Time and Space Complexity

The given Python code defines a function constrainedSubsetSum which calculates the maximum sum of a non-empty subsequence of the array nums, where the subsequence satisfies the constraint that no two elements are farther apart than k in the original array.

Time Complexity

The time complexity of the code is O(n), where n is the number of elements in the array nums. This is because the algorithm iterates over each element exactly once. Inside the loop, it performs operations that are constant time on average, such as adding or removing elements from the deque q. The deque maintains the maximum sum at each position within the window of k, and the operations are done in constant time due to the nature of the double-ended queue.

Even though we have a while-loop inside the for-loop that pops elements from the deque, this does not increase the overall time complexity. Each element is added once and removed at most once from the deque, leading to an average constant time for these operations per element.

Space Complexity

The space complexity of the code is O(n), where n is the size of nums. This is because the algorithm allocates an array dp of the same size as nums to store the maximum sum up to each index. Furthermore, the deque q can at most contain k elements, where k is the constraint on the distance between the elements in the subsequence. However, since k is a constant with respect to n, the primary factor that affects the space complexity is the dp array.

In summary:

• Time complexity is O(n).
• Space complexity is O(n).

Learn more about how to find time and space complexity quickly using problem constraints.