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)
class Solution {
public:
    vector<int> topKFrequent(vector<int>& nums, int k) {
        unordered_map<int, int> cnt;
        using pii = pair<int, int>;
        for (int x : nums) {
            ++cnt[x];
        }
        priority_queue<pii, vector<pii>, greater<pii>> pq;
        for (auto& [x, c] : cnt) {
            pq.push({c, x});
            if (pq.size() > k) {
                pq.pop();
            }
        }
        vector<int> ans;
        while (!pq.empty()) {
            ans.push_back(pq.top().second);
            pq.pop();
        }
        return ans;
    }
};
