# 692 Top K Frequent Words

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?

 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 `````` ``````class Pair: def __init__(self, word, freq): self.word = word self.freq = freq def __lt__(self, p): return ( self.freq < p.freq or (self.freq == p.freq and self.word > p.word) ) class Solution: def topKFrequent(self, words: List[str], k: int) -> List[str]: heap = [] counts = defaultdict(int) for word in words: counts[word] += 1 for word, count in counts.items(): heapq.heappush(heap, Pair(word, count)) if len(heap) > k: heapq.heappop(heap) return [p.word for p in sorted(heap, reverse=True)]``````