Problem URL: Top k frequent words

First we will count all the words and then create a max heap with the count, each element will have 2 items, first will be the count, second will be the word itself. Then we pop k elements and put the words in a list and return it.

class Solution:
    def topKFrequent(self, words: List[str], k: int) -> List[str]:
        count = [(-t, w) for w, t in collections.Counter(words).items()]
        heapq.heapify(count)

        return [heapq.heappop(count)[1] for _ in range(k)]

Time Complexity: O(n + k * log(n)), n in the the number of words, k is the top elements.
Space Complexity: O(n)