# 692. Top K Frequent Words

#### Medium

***

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:**

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

**Example 2:**

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

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**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]]`

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

```python
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]
```