<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>

import heapq
class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        #Put them into a dictionary
        #Put them into a heap
        #Pop from heap until there is k left or pop k from the heap.
        numsdict = {}
        heap = []
        for i in nums:
            if i in numsdict:
                numsdict[i] += 1
            else:
                numsdict[i] = 1
        
        for key, value in numsdict.items():
            index = (value, key)
            heap.append(index)
        
        heapq.heapify(heap)
        while len(heap) > k:
            heapq.heappop(heap)

        answer = []
        while heap:
            answer.append( heapq.heappop(heap)[1])    
        return answer    
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