Cap Repeated Values in a Sorted Array

# Cap Repeated Values in a Sorted Array You are given an integer array `nums`, sorted in nondecreasing order, and an integer `k`. Return the array obtained by limiting each distinct value to at most `k` occurrences while preserving order. If a value occurs at least `k` times, retain exactly `k` copies. If it occurs fewer than `k` times, retain all its copies. ### Function `vector<int> limitOccurrences(vector<int>& nums, int k)` ### Example 1 **Input:** `nums = [2, 2, 2, 2, 5, 5, 8], k = 3` **Output:** `[2, 2, 2, 5, 5, 8]` **Explanation:** Retain three of the four copies of `2`, both copies of `5`, and the single `8`. Their order is unchanged. ### Example 2 **Input:** `nums = [4, 4, 6, 6, 6, 9, 9], k = 1` **Output:** `[4, 6, 9]` **Explanation:** Each distinct value is retained once. ### Constraints - `1 <= nums.length <= 100` - `1 <= nums[i] <= 100` - `nums` is sorted in nondecreasing order. - `1 <= k <= nums.length` ### Clarifications - Return the resulting values, not just their count. - Values already appearing at most `k` times are retained without change. - The array is already sorted; no rearrangement is needed or allowed in the result. ### What a Strong Answer Covers - Retaining the required number of occurrences from every value, including the first and last runs. - Preserving order and handling `k = 1` as well as limits that remove nothing. - Returning the complete resulting array through the specified interface. ### Follow-up Questions - Can you do the filtering in place with `O(1)` extra space? Space used for returning or resizing the result does not count toward that extra-space requirement.