692. Top K Frequent Words

Medium

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Given an array of strings words and an integer k, return the k most frequent strings.

Return the answer sorted by the frequency from highest to lowest. Sort the words with the same frequency by their lexicographical order.

Example 1:

Input: words = ["i","love","leetcode","i","love","coding"], k = 2
Output:
 ["i","love"]
Explanation:
 "i" and "love" are the two most frequent words.
Note that "i" comes before "love" due to a lower alphabetical order.

Example 2:

Input: words = ["the","day","is","sunny","the","the","the","sunny","is","is"], k = 4
Output:
 ["the","is","sunny","day"]
Explanation:
 "the", "is", "sunny" and "day" are the four most frequent words, with the number of occurrence being 4, 3, 2 and 1 respectively.

Constraints:

• 1 <= words.length <= 500

• 1 <= words[i].length <= 10

• words[i] consists of lowercase English letters.

• k is in the range [1, The number of unique words[i]]

Follow-up: Could you solve it in O(n log(k)) time and O(n) extra space?

class Word:
    def __init__(self, word, count):
        self.word = word
        self.count = count
    def __lt__(self, other):
        if self.count == other.count:
            return self.word > other.word
        return self.count < other.count
    def __eq__(self, other):
        return self.count == other.count and self.word == other.word

class Solution:
    def topKFrequent(self, words: List[str], k: int) -> List[str]:
        counter = Counter(words)
        h = []
        for word, count in counter.items():
            heapq.heappush(h, Word(word, count))
            if len(h) > k:
                heapq.heappop(h)
        result = []
        for _ in range(k):
            result.append(heapq.heappop(h).word)
        return result[::-1]