Description
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.
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
Solution
Python3
class Solution:
def constrainedSubsetSum(self, nums: List[int], k: int) -> int:
N = len(nums)
queue = deque()
res = -inf
# dp[i] = nums[i] + max(0, dp[i - 1], dp[i - 2], ..., dp[i - k])
# O(N) solution: keep a decreasing monotonic queue
for i, x in enumerate(nums):
curr = x
if queue:
curr += max(0, queue[0][0])
while queue and curr > queue[-1][0]:
queue.pop()
if curr > 0:
queue.append((curr, i))
if queue and i - queue[0][1] == k:
queue.popleft()
res = max(res, curr)
return res