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Constrained Subsequence Sum

Hard
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Topics:
ArraysDynamic ProgrammingGreedy Algorithms

Given an integer array nums and an integer k, return the maximum sum of a non-empty subsequence of that array such that for every two consecutive integers in the subsequence, nums[i] and nums[j], where i < j, the condition j - i <= k is satisfied.

A subsequence of an array is obtained by deleting some number of elements (can be zero) from the array, leaving the remaining elements in their original order.

For example:

  • nums = [10,2,-10,5,20], k = 2. The output should be 37 because the subsequence is [10, 2, 5, 20].
  • nums = [-1,-2,-3], k = 1. The output should be -1 because the subsequence must be non-empty, so we choose the largest number.
  • nums = [10,-2,-10,-5,20], k = 2. The output should be 23 because the subsequence is [10, -2, -5, 20].

Explain your approach and provide code.

Solution


Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

  1. What are the constraints on the values within the input array `nums`? Can they be negative, zero, or floating-point numbers?
  2. What is the maximum size of the input array `nums`? I want to ensure my solution is efficient.
  3. If no subsequence satisfies the constraint `k`, should I return 0, null, or some other specific value?
  4. Could you clarify what is expected when there are multiple subsequences that have the same maximum sum? Should I return any of them, or is there a specific subsequence that is preferred?
  5. Is the input `k` always a non-negative integer, or can it be negative or zero?

Brute Force Solution

Approach

The brute force approach to finding the best subsequence sum with constraints involves checking every possible subsequence. We look at each potential subsequence, make sure it meets the distance requirement between elements, and calculate its sum. Finally, we compare all the sums to find the largest one.

Here's how the algorithm would work step-by-step:

  1. Start with the first number in the sequence.
  2. Consider this number alone as a potential subsequence, and remember its value.
  3. Now, add the next number to the sequence, but only if it's close enough to the previous number (satisfies the distance constraint).
  4. If it meets the distance requirement, include the number in the subsequence and update the running sum.
  5. Continue adding numbers as long as they meet the distance requirement.
  6. Record the sum of this subsequence.
  7. Repeat this process, starting with the first number, but this time excluding different numbers from the sequence to create different subsequences.
  8. Do this for every possible starting point in the original sequence and for every possible combination of numbers that still fit the distance requirement.
  9. After checking every possible subsequence and calculating their sums, find the largest sum among all the valid subsequences.

Code Implementation

def constrained_subsequence_sum_brute_force(numbers, constraint):
    max_sum_found = float('-inf')
    list_length = len(numbers)

    for start_index in range(list_length):
        # Iterate over each number in the list as the starting point
        for i in range(1 << list_length):
            current_subsequence = []
            current_sum = 0
            last_included_index = -1

            for j in range(list_length):
                # Check if the j-th element is included in the current subsequence
                if (i >> j) & 1:
                    # Ensure that the distance constraint is met
                    if last_included_index == -1 or j - last_included_index <= constraint:
                        current_subsequence.append(numbers[j])
                        current_sum += numbers[j]
                        last_included_index = j
                    else:
                        # If the constraint is violated, break the subsequence
                        current_sum = float('-inf')
                        break

            # Update the maximum sum if the current subsequence sum is larger
            max_sum_found = max(max_sum_found, current_sum)

    return max_sum_found

Big(O) Analysis

Time Complexity
O(2^n)The provided brute force approach examines all possible subsequences of the input array. For each element, we have a choice: either include it in the subsequence or exclude it. With n elements, this results in 2^n possible subsequences. Therefore, the time complexity is O(2^n) because we potentially evaluate the sum and validity of each of these subsequences.
Space Complexity
O(1)The brute force approach, as described, primarily involves iterating through the input array and calculating sums. It doesn't explicitly mention the creation of auxiliary data structures like lists, maps, or arrays to store intermediate subsequences. While it implicitly maintains a variable to store the current subsequence's sum and a variable to store the maximum sum found so far, these are constant space operations. Therefore, the space complexity remains constant regardless of the input array size N.

Optimal Solution

Approach

The best way to find the biggest subsequence sum with distance constraints is to keep track of useful sums along the way. It involves intelligently maintaining a pool of the most promising subsequences, discarding the ones that won't lead to an optimal solution.

Here's how the algorithm would work step-by-step:

  1. Imagine you're walking through a list of numbers, one by one.
  2. As you go, calculate the best possible sum you can get by including the current number in a subsequence.
  3. To figure this out, consider the best sums you've seen recently (within the allowed distance) and add the current number to them.
  4. Keep only the 'good' sums. A sum is 'good' if it's bigger than anything you've seen so far, or if it's the biggest among the recent sums.
  5. If a sum is too small or too far in the past, it's no longer useful, so discard it.
  6. At the end, the biggest sum you've kept track of is the answer.

Code Implementation

from collections import deque

def constrained_subset_sum(numbers, allowed_distance):
    maximum_sum = float('-inf')
    monotonically_decreasing_queue = deque()

    for i, number in enumerate(numbers):
        # The current element can always start a new subsequence.
        current_sum = number

        # Check if there's a valid previous subsequence to extend.
        if monotonically_decreasing_queue:
            current_sum = max(current_sum, number + monotonically_decreasing_queue[0])

        maximum_sum = max(maximum_sum, current_sum)

        # Only store positive sums as they can contribute to larger subsequences.
        if current_sum > 0:
            while monotonically_decreasing_queue and current_sum >= monotonically_decreasing_queue[-1]:
                monotonically_decreasing_queue.pop()
            monotonically_decreasing_queue.append(current_sum)

        # Remove elements outside the allowed distance.
        while monotonically_decreasing_queue and i - monotonically_decreasing_queue[-1] > allowed_distance:
            monotonically_decreasing_queue.popleft()

        # The queue stores index of element, not the element itself.
        monotonically_decreasing_queue = deque([x for x in monotonically_decreasing_queue if i - numbers.index(number) <= allowed_distance])

    return maximum_sum

Big(O) Analysis

Time Complexity
O(n)The algorithm iterates through the array of n numbers once. Inside the loop, a monotonic queue (deque) is used to store the indices of potentially optimal subsequence sums within the constraint k. Adding and removing elements from the monotonic queue takes O(1) time on average. Therefore, the dominant operation is iterating through the array once, resulting in a time complexity of O(n).
Space Complexity
O(K)The algorithm maintains a pool of 'good' sums within the allowed distance K. This pool can be implemented using a data structure such as a deque or a list of size at most K to store these sums. Therefore, the auxiliary space required is proportional to the constraint K, used to keep track of the most promising subsequences within the specified distance. Consequently, the space complexity is O(K).

Edge Cases

CaseHow to Handle
Empty input arrayReturn 0 since there are no subsequences.
Array with a single elementReturn the single element if it's non-negative, otherwise return 0.
All elements are negativeReturn 0 since the maximum subsequence sum will be 0 (empty subsequence).
Large input array (performance considerations)Use a Deque (double-ended queue) for efficient maintenance of the maximum value within the k window to ensure the solution scales linearly O(n).
k = 0If k is 0, it means there is no constraint on the distance between elements, thus we can pick all positive numbers.
k is greater than or equal to the length of the arrayEffectively, there's no constraint, so the subsequence can be any combination of elements.
Integer overflow when calculating subsequence sumUse long data type to store the intermediate sums to prevent overflow, and ensure to cast array values to long when needed.
Array contains very large positive and negative numbersThe dynamic programming and deque approach handles these cases correctly as it tracks the maximum sum within the constrained window.