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.
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:
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:
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
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:
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
Case | How to Handle |
---|---|
Empty input array | Return 0 since there are no subsequences. |
Array with a single element | Return the single element if it's non-negative, otherwise return 0. |
All elements are negative | Return 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 = 0 | If 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 array | Effectively, there's no constraint, so the subsequence can be any combination of elements. |
Integer overflow when calculating subsequence sum | Use 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 numbers | The dynamic programming and deque approach handles these cases correctly as it tracks the maximum sum within the constrained window. |