<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>

class Solution:
    def topKFrequent(self, words: List[str], k: int) -> List[str]:
        word_to_freq = {}
        for word in words:
            if word in word_to_freq:
                word_to_freq[word] += 1
            else:
                word_to_freq[word] = 1
        freq_to_word = {}
        for word in word_to_freq:
            freq = word_to_freq[word]
            if freq in freq_to_word:
                freq_to_word[freq].append(word)
            else:
                freq_to_word[freq] = [word]
        #sorted_freq_to_word = {key: freq_to_word[key] for key in sorted(freq_to_word.keys(), reverse=True)}
        sorted_freq_to_word = {key: value for key, value in sorted(freq_to_word.items(), key=lambda item: item[0], reverse=True)}
        for key in sorted_freq_to_word:
            sorted_freq_to_word[key] = sorted(sorted_freq_to_word[key])
        solutions = []
        for key in sorted_freq_to_word:
            if k <= 0:
                break
            if len(sorted_freq_to_word[key]) <= k:
                solutions += sorted_freq_to_word[key]
                k -= len(sorted_freq_to_word[key])
            else:
                solutions += sorted_freq_to_word[key][:k]
                k -= k
        return solutions

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