Problem URL: Top k frequent element

We will count the frequency of each element and store it in a hashmap, where the number itself will be the key and frequency will be the value. Then we sort the elements of the hashmap and get the keys as a list. Then we return first k elements of the list.

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        count = {}
        for n in nums:
            count[n] = 1 + count.get(n, 0)
        count = dict(sorted(count.items(), key=lambda item: -item[1]))
        return list(count.keys())[:k]

We sort the elements of the hashmap, in worst case that is O(nlog(n)) time complexity. We also store the frequency of the list, that is O(n) space complexity.