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Java-based LeetCode algorithm problem solutions, regularly updated
2382\. Maximum Segment Sum After Removals
Hard
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**.