Top K Frequent Words
<p>Given an array of strings <code>words</code> and an integer <code>k</code>, return <em>the </em><code>k</code><em> most frequent strings</em>.</p>

<p>Return the answer <strong>sorted</strong> by <strong>the frequency</strong> from highest to lowest. Sort the words with the same frequency by their <strong>lexicographical order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

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

<p><strong class="example">Example 2:</strong></p>

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

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= words.length &lt;= 500</code></li>
	<li><code>1 &lt;= words[i].length &lt;= 10</code></li>
	<li><code>words[i]</code> consists of lowercase English letters.</li>
	<li><code>k</code> is in the range <code>[1, The number of <strong>unique</strong> words[i]]</code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow-up:</strong> Could you solve it in <code>O(n log(k))</code> time and <code>O(n)</code> extra space?</p>
Trie; Array; Hash Table; String; Bucket Sort; Counting; Sorting; Heap (Priority Queue)
class Solution:
    def topKFrequent(self, words: List[str], k: int) -> List[str]:
        cnt = Counter(words)
        return sorted(cnt, key=lambda x: (-cnt[x], x))[:k]
