Return the k most frequent values in an array with a deterministic tie rule

Given an integer array `nums` and an integer `k`, return the `k` values that occur most often in `nums`. Rank the distinct values of `nums` by frequency (number of occurrences), highest first, and break ties between values with equal frequency by putting the smaller value first. Return the first `k` values of this ranking, in ranking order. ### Function Signature ```python def k_most_frequent(nums: list[int], k: int) -> list[int]: ``` ### Rules - Each distinct value appears at most once in the output. - The output order is the ranking order: frequency descending, then value ascending. - The same rule decides which values are included when several values tie at the cutoff: a smaller value always ranks ahead of a larger value with the same frequency. ### Constraints - `1 <= len(nums) <= 10^5` - `-10^9 <= nums[i] <= 10^9` - `1 <= k <= d`, where `d` is the number of distinct values in `nums` ### Examples **Example 1** ```text Input: nums = [4, 1, 4, 2, 1, 4, 3], k = 2 Output: [4, 1] ``` `4` occurs three times, `1` twice, and `2` and `3` once each. **Example 2** ```text Input: nums = [3, 3, 2, 2, 1, 1, 0], k = 2 Output: [1, 2] ``` `1`, `2` and `3` each occur twice. The tie is broken by the smaller value, so `1` and `2` are returned and `3` is not. **Example 3** ```text Input: nums = [-5, 7, -5, 7, 9], k = 3 Output: [-5, 7, 9] ``` `-5` and `7` each occur twice, and `-5` is smaller. `9` occurs once.