Top K Frequent Elements
<p>Given an integer array <code>nums</code> and an integer <code>k</code>, return <em>the</em> <code>k</code> <em>most frequent elements</em>. You may return the answer in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> nums = [1,1,1,2,2,3], k = 2
<strong>Output:</strong> [1,2]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> nums = [1], k = 1
<strong>Output:</strong> [1]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
	<li><code>k</code> is in the range <code>[1, the number of unique elements in the array]</code>.</li>
	<li>It is <strong>guaranteed</strong> that the answer is <strong>unique</strong>.</li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Your algorithm&#39;s time complexity must be better than <code>O(n log n)</code>, where n is the array&#39;s size.</p>
Array; Hash Table; Divide and Conquer; Bucket Sort; Counting; Quickselect; Sorting; Heap (Priority Queue)
function topKFrequent(nums: number[], k: number): number[] {
    const cnt = new Map<number, number>();
    for (const x of nums) {
        cnt.set(x, (cnt.get(x) ?? 0) + 1);
    }
    const pq = new MinPriorityQueue();
    for (const [x, c] of cnt) {
        pq.enqueue(x, c);
        if (pq.size() > k) {
            pq.dequeue();
        }
    }
    return pq.toArray().map(x => x.element);
}
