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Kotlin-based LeetCode algorithm problem solutions, regularly updated
2454\. Next Greater Element IV
Hard
You are given a **0-indexed** array of non-negative integers `nums`. For each integer in `nums`, you must find its respective **second greater** integer.
The **second greater** integer of `nums[i]` is `nums[j]` such that:
* `j > i`
* `nums[j] > nums[i]`
* There exists **exactly one** index `k` such that `nums[k] > nums[i]` and `i < k < j`.
If there is no such `nums[j]`, the second greater integer is considered to be `-1`.
* For example, in the array `[1, 2, 4, 3]`, the second greater integer of `1` is `4`, `2` is `3`, and that of `3` and `4` is `-1`.
Return _an integer array_ `answer`_, where_ `answer[i]` _is the second greater integer of_ `nums[i]`_._
**Example 1:**
**Input:** nums = [2,4,0,9,6]
**Output:** [9,6,6,-1,-1]
**Explanation:**
0th index: 4 is the first integer greater than 2, and 9 is the second integer greater than 2, to the right of 2.
1st index: 9 is the first, and 6 is the second integer greater than 4, to the right of 4.
2nd index: 9 is the first, and 6 is the second integer greater than 0, to the right of 0.
3rd index: There is no integer greater than 9 to its right, so the second greater integer is considered to be -1.
4th index: There is no integer greater than 6 to its right, so the second greater integer is considered to be -1.
Thus, we return [9,6,6,-1,-1].
**Example 2:**
**Input:** nums = [3,3]
**Output:** [-1,-1]
**Explanation:** We return [-1,-1] since neither integer has any integer greater than it.
**Constraints:**
* 1 <= nums.length <= 105
* 0 <= nums[i] <= 109