<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]:
      count = {}
      freq = [[] for i in range(len(nums)+1)]

      for n in nums:
        count[n] = 1 + count.get(n,0)
      for n,c in count.items():
        freq[c].append(n)

      res = []
      for i in range(len(freq)-1,0,-1):
        for n in freq[i]:
          res.append(n)
          if len(res) == k:
            return res

      #My Solution     
      # new_arr = set(nums)
      # dic = {}
      # for i in new_arr:
      #   dic[i] = nums.count(i)
      # sorted_dic = dict(sorted(dic.items() , key = lambda item : item[1] , reverse=True))
      # res = list(sorted_dic.keys())
      # return res[:k]


        
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