Return the K Most Frequent Values with a Larger-Value Tie-Break

## Problem Given an integer array and an integer `k`, return the `k` distinct values with the highest frequencies. Sort the result by descending frequency. If two values have equal frequency, place the larger value first. ### Function Contract Implement `topKFrequentWithTieBreak(values, k)` and return an integer array. ### Constraints & Assumptions - `1 <= len(values) <= 200,000`. - Values are integers in the signed 32-bit range. - `1 <= k <= number of distinct values`. - The output order is part of the contract. ### Clarifying Questions to Ask - Are duplicate values counted separately? Yes, each occurrence contributes to frequency. - What resolves a frequency tie? The numerically larger value comes first. - Must the full set of distinct values be sorted? No; an efficient bounded selection is acceptable. ```hint Give the heap the inverse priority When retaining only `k` candidates, the root should be the worst retained candidate: lower frequency first, and for an equal frequency, smaller value first. ``` ### Examples ```text values = [1,1,1,2,2,3], k = 2 -> [1,2] values = [4,4,3,3,2], k = 2 -> [4,3] values = [-1,-1,2,2], k = 2 -> [2,-1] ``` ### Evaluation Focus - Counts every distinct value accurately. - Applies the larger-value tie-break both during selection and final ordering. - Returns exactly `k` values. - Achieves `O(n + m log k)` time for `m` distinct values, or justifies another efficient bound. ### Extensions to Discuss 1. How would you maintain the answer over a continuous event stream? 2. What changes if the tie-break is earliest first occurrence? 3. When is bucket sorting preferable to a heap?