<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>

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        counts = {}
        vals = []
        for num in nums:
            if num not in counts:
                vals.append(num)
                counts[num] = 0
            counts[num] += 1
        n = len(vals)
        k = n - k
        return self.quickselect(vals, 0, n - 1, k, counts)
    
    def quickselect(self, vals, left, right, k, counts):
        pivot_idx = randint(left, right)
        pivot_val = vals[pivot_idx]
        pivot_count = counts[pivot_val]
        cur_idx = left
        self.swap(vals, pivot_idx, right)
        for i in range(left, right):
            val = vals[i]
            val_count = counts[val]
            if val_count < pivot_count:
                self.swap(vals, cur_idx, i)
                cur_idx += 1
        self.swap(vals, cur_idx, right)
        if cur_idx < k:
            return self.quickselect(vals, cur_idx + 1, right, k, counts)
        elif k < cur_idx:
            return self.quickselect(vals, left, cur_idx - 1, k, counts)
        return vals[k:]

    def swap(self, vals, i, j):
        vals[i], vals[j] = vals[j], vals[i]

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