{leetcode}/problems/maximum-segment-sum-after-removals/[LeetCode - 2382. Maximum Segment Sum After Removals ^]
You are given two 0-indexed integer arrays nums
and removeQueries
, both of length n
. For the ith
query, the element in nums
at the index removeQueries[i]
is removed, splitting nums
into different segments.
A segment is a contiguous sequence of positive integers in nums
. A segment sum is the sum of every element in a segment.
Return_ an integer array `answer`, of length `n`, where `answer[i] is the maximum segment sum after applying the `ith
_removal.
Note: The same index will not be removed more than once.
Example 1:
Input: nums = [1,2,5,6,1], removeQueries = [0,3,2,4,1] Output: [14,7,2,2,0] Explanation: Using 0 to indicate a removed element, the answer is as follows: Query 1: Remove the 0th element, nums becomes [0,2,5,6,1] and the maximum segment sum is 14 for segment [2,5,6,1]. Query 2: Remove the 3rd element, nums becomes [0,2,5,0,1] and the maximum segment sum is 7 for segment [2,5]. Query 3: Remove the 2nd element, nums becomes [0,2,0,0,1] and the maximum segment sum is 2 for segment [2]. Query 4: Remove the 4th element, nums becomes [0,2,0,0,0] and the maximum segment sum is 2 for segment [2]. Query 5: Remove the 1st element, nums becomes [0,0,0,0,0] and the maximum segment sum is 0, since there are no segments. Finally, we return [14,7,2,2,0].
Example 2:
Input: nums = [3,2,11,1], removeQueries = [3,2,1,0] Output: [16,5,3,0] Explanation: Using 0 to indicate a removed element, the answer is as follows: Query 1: Remove the 3rd element, nums becomes [3,2,11,0] and the maximum segment sum is 16 for segment [3,2,11]. Query 2: Remove the 2nd element, nums becomes [3,2,0,0] and the maximum segment sum is 5 for segment [3,2]. Query 3: Remove the 1st element, nums becomes [3,0,0,0] and the maximum segment sum is 3 for segment [3]. Query 4: Remove the 0th element, nums becomes [0,0,0,0] and the maximum segment sum is 0, since there are no segments. Finally, we return [16,5,3,0].
Constraints:
-
n == nums.length == removeQueries.length
-
1 ⇐ n ⇐ 105
-
1 ⇐ nums[i] ⇐ 109
-
0 ⇐ removeQueries[i] < n
-
All the values of
removeQueries
are unique.