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>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [1,1,1,2,2,3], k = 2</span></p>

<p><strong>Output:</strong> <span class="example-io">[1,2]</span></p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [1], k = 1</span></p>

<p><strong>Output:</strong> <span class="example-io">[1]</span></p>
</div>

<p><strong class="example">Example 3:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [1,2,1,2,1,2,3,1,3,2], k = 2</span></p>

<p><strong>Output:</strong> <span class="example-io">[1,2]</span></p>
</div>

<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)
use std::cmp::Reverse;
use std::collections::{BinaryHeap, HashMap};

impl Solution {
    pub fn top_k_frequent(nums: Vec<i32>, k: i32) -> Vec<i32> {
        let mut cnt = HashMap::new();
        for x in nums {
            *cnt.entry(x).or_insert(0) += 1;
        }
        let mut pq = BinaryHeap::with_capacity(k as usize);
        for (&x, &c) in cnt.iter() {
            pq.push(Reverse((c, x)));
            if pq.len() > k as usize {
                pq.pop();
            }
        }
        pq.into_iter().map(|Reverse((_, x))| x).collect()
    }
}
