Majority Element II

https://leetcode.com/problems/majority-element-ii/description/

class Solution:
    def majorityElement(self, nums: List[int]) -> List[int]:
        count = defaultdict(int)

        for n in nums:
            count[n] += 1

            if len(count) <= 2:
                continue

            new_count = defaultdict(int)
            for n, c in count.items():
                if c > 1:
                    new_count[n] = c - 1
            count = new_count

        res = []
        for n in count:
            if nums.count(n) > len(nums) // 3:
                res.append(n)
        return res

we cannot have more than 2 answer since we are trying to find numbers that occur more than n / 3.

in hashmap we keep put numbers and count occurence.

when we have 3rd entry we decrement by 1 and remove if occurence become zero.

and later on check the numbers that are still in hashmap.