• Home
• com.github.javadev
• leetcode-in-kotlin
• 1.21
• source code
• readme.md

×

Project download

Many resources are needed to download a project. Please understand that we have to compensate our server costs. Thank you in advance.
Project price only 1 $

You can buy this project and download/modify it how often you want.

g2301_2400.s2382_maximum_segment_sum_after_removals.readme.md Maven / Gradle / Ivy

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**.