692-top-k-frequent-words

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Description

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] <= 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?

Solution(Python)

class Comparator:
    def __init__(self, count, word):
        self.count = count
        self.word = word

    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]:
        return self.priorityqueue(words, k)

    # Time Complexity: O(nmlogn)
    # Space Compexlity: O(n*m)
    def sorting(self, words: List[str], k: int) -> List[str]:
        counter = Counter(words)
        counter = sorted(counter.keys(), key=counter.get, reverse=True)
        return counter[:k]

    # Time Complexity: O(nmlogk)
    # Space Compexlity: O(k)
    def priorityqueue(self, words: List[str], k: int) -> List[str]:
        heap = []
        counter = Counter(words)
        heapq.heapify(heap)
        for word in counter:
            heapq.heappush(heap, (Comparator(counter[word], word), word))
            if len(heap) > k:
                heapq.heappop(heap)

        res = []
        for _ in range(k):
            res.append(heapq.heappop(heap)[1])
        return res[::-1]