'''
Given a non-empty array of integers, return the k most frequent elements.

For example,
Given [1,1,1,2,2,3] and k = 2, return [1,2].

Note:
You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
Your algorithm's time complexity must be better than O(n log n), where n is the array's size.
'''
from collections import Counter

class Solution:
    def topKFrequent(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: List[int]
        """
        dic = [(n, c) for n, c in Counter(nums).items()]
        dic.sort(key=lambda x: -x[1])

        print([n[0] for n in dic[:k]])


s = Solution()

nums = [1,1,1,2,2,3]
k = 2

s.topKFrequent(nums, k )

