Kth Greater Element Indexes

You are given an integer array nums and an integer k. For each index i, look at the elements to the right of i whose value is strictly greater than nums[i], preserving their original left-to-right order.
Return an integer array answer where answer[i] is the 1-indexed position of the k-th such greater element for index i. If fewer than k greater elements appear to the right of i, set answer[i] to -1.

Function
kthGreaterElementIndexes(nums: int[], k: int) → int[]

Examples
Example 1
nums = [3, 4, 2, 6, 5]
k = 2
return = [4, 5, 5, -1, -1]
For 3, the greater elements to the right are 4, 6, 5, so the 2nd one is 6 at 1-indexed position 4. For 4, the greater elements are 6, 5, so the answer is position 5.
Example 2
nums = [5, 1, 4, 2, 3]
k = 1
return = [-1, 3, -1, 5, -1]
For 1, the first greater element to its right is 4 at position 3. For 2, it is 3 at position 5.

Constraints
1 <= nums.length <= 10^5
1 <= k <= nums.length
-10^9 <= nums[i] <= 10^9

Brute force is O(n^2) and times out at 10^5. The trick is to process values in descending order and keep a sorted structure of positions. Sort indices by value, and for each group of equal values, query before inserting them, since strictly greater means equal values must not count. For index i, you need the k-th position greater than i among inserted positions. Use a Fenwick tree over positions with count prefix sums, then binary-search descent to find the (rank of i + k)-th inserted position. Rank is the count of inserted positions up to i. If total inserted is below that, answer is -1. The common pitfall is inserting equal values before querying, which wrongly counts ties. Another is forgetting the answer is 1-indexed. If the Fenwick descent feels shaky under pressure, StealthCoder is the hedge for the live OA, since it reads the problem and hands you a working solution.

Memorize the pattern. If you can't, run StealthCoder. The proctor sees the IDE. They don't see what's behind it.