Top K Frequent Elements

Given an integer array nums and an integer k, return the k distinct values with the highest frequencies.
Order the answer by decreasing frequency. If two selected values have equal frequency, place the smaller value first.

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

Examples
Example 1
nums = [1,1,1,2,2,3]
k = 2
return = [1,2]
1 occurs three times and 2 occurs twice.
Example 2
nums = [4,4,-1,-1,2]
k = 2
return = [-1,4]
The selected values tie in frequency, so the smaller value comes first.

Constraints
1 <= nums.length <= 100000.
-10^9 <= nums[i] <= 10^9.
1 <= k <= the number of distinct values.

Step one: count frequencies with a hash map. Step two: rank the distinct values by frequency descending, then value ascending. You can sort the distinct entries with that comparator in O(d log d), or use a min-heap of size k with the same ordering to get O(d log k). Either passes at 100000 elements. The pitfall is the tie-break. Classic solutions skip it, and Example 2 exists to catch you: -1 and 4 both appear twice, so -1 goes first. Another trap is negative values, so don't use a bucket array indexed by the number itself. Bucket by frequency instead if you want O(n), then sort within each bucket. If you freeze on the comparator during the live OA, StealthCoder runs invisibly and can hand you the working version. Test your tie case before submitting.

If you see this problem in your OA tomorrow, the play is to recognize the pattern in 30 seconds. StealthCoder buys you that recognition.