Day-17_Top_K_Frequent_Elements.py

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

Example 1:

Input: nums = [1,1,1,2,2,3], k = 2
Output: [1,2]
Example 2:

Input: nums = [1], k = 1
Output: [1]
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.
It's guaranteed that the answer is unique, in other words the set of the top k frequent elements is unique.
You can return the answer in any order.


'''

class Solution:
    def topKFrequent(self, nums, k):
        bucket = [[] for _ in range(len(nums) + 1)]
        Count = Counter(nums).items()  
        for num, freq in Count: bucket[freq].append(num) 
        flat_list = list(chain(*bucket))
        return flat_list[::-1][:k]