Problem
Given a nonempty 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.
Thought Process and White Board
 Use a hashtable to count the frequency of each number
 Sort the hashtable
 Return the top k elements by value
Code
class Solution:
def _create_ht(self, nums):
ht = {}
for num in nums:
if num in ht:
ht[num] += 1
else:
ht[num] = 1
return ht
def _ord_dict_val(self, my_dict, k):
ord_tuples = sorted([(v,k) for (k,v) in my_dict.items()], reverse=True)[:k]
return {k:v for (v,k) in ord_tuples}
def topKFrequent(self, nums, k):
ht = self._create_ht(nums)
if k < 0 or k > len(ht):
return 1
ord_dict = self._ord_dict_val(ht, k)
return ord_dict.keys()
Optimize

As shown in the output, the submission was accepted but it's outperformed by most submissions. The reason is because sorting has a complexity of O(nlogn). An improved solution would be:
 create a priority queue of k elements
 if the current element has a higher count than the min element in the queue, then replace the min element in the queue with the current element
 finally, the queue will have the top k freq elements
 The complexity is O(nlogk), n is len(nums), k is the count of result, it would be smaller than O(nlogn)