<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]:
        counter = defaultdict(list)
        final_list = []
        remaining = k
        temp_set = set(nums)
        if k == len(nums):
            return nums
        for i in temp_set:
            count = nums.count(i)
            counter[count].append(i)
        while len(final_list) < k:
            max_count = max(counter.keys())
            # Append the numbers with the highest frequency
            final_list.extend(counter[max_count])
            counter.pop(max_count)
        
        # Ensure only top k elements are returned
        return final_list[:k]
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