Elements Occurring More Than n/3 Times

Given an integer array nums of length n, return every distinct value whose frequency is strictly greater than n / 3.
Return qualifying values in increasing numerical order. Frequency counts occurrences of the whole integer value, rather than its individual decimal digits. Return an empty array if no value qualifies.

Function
majorityOverThird(nums: int[]) → int[]

Examples
Example 1
nums = [3,2,3]
return = [3]
The value 3 occurs twice, strictly more than 3 / 3 = 1.
Example 2
nums = [1,2,1,2,1,2,3]
return = [1,2]
Both 1 and 2 occur three times, more than 7 / 3.
Example 3
nums = [1,2,3,4]
return = []
No frequency exceeds 4 / 3.

Constraints
0 <= nums.length <= 10^5
-10^9 <= nums[i] <= 10^9
For integer frequencies, the threshold is equivalent to frequency > floor(n / 3).

The trick is that no more than two distinct values can each exceed n/3, since three of them would need more than n elements. That unlocks Boyer-Moore voting with two candidates and two counters. Pass one: match a candidate, else fill an empty slot, else decrement both counters. Pass two: count the real frequencies of both candidates and keep those where count > floor(n/3). The second pass is the one people skip, and it's the common pitfall, because the survivors of pass one aren't guaranteed to qualify. Also sort the result before returning, since the output must be increasing. Handle n = 0 by returning an empty array. The hash map version is O(n) time and O(n) space, and it's perfectly acceptable if the OA only checks correctness. If you freeze mid-assessment, StealthCoder can surface the two-candidate voting code so you can verify it against the three examples.

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