Top K Frequent Elements

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.

solution is similar to bucket sort.  with help or hash table, this can be O(n).
1. calculate statistics of frequency by using hash table/map
2. putting map entries to priority queue by defining proper comparator that makes it sorting by frequency high to low

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public class Solution { public List<Integer> topKFrequent(int[] nums, int k) { HashMap<Integer,Integer> frequencyMap = new HashMap<Integer,Integer>(); //counte frequence with dict/hashmap/hashtable for(int i=0;i<nums.length;i++){ if(frequencyMap.get(nums[i])==null){ frequencyMap.put(nums[i],1); } else{ frequencyMap.put(nums[i],frequencyMap.get(nums[i])+1); } } //sort according to frequency by putting them in priority queue backed by reversed number order PriorityQueue<Map.Entry<Integer, Integer>> pq= new PriorityQueue<Map.Entry<Integer, Integer>>( new Comparator<Map.Entry<Integer, Integer>>(){ public int compare(Map.Entry<Integer, Integer> e1,Map.Entry<Integer, Integer> e2){ return e2.getValue()-e1.getValue(); } }); pq.addAll(frequencyMap.entrySet()); List<Integer> aList = new ArrayList<Integer>(); //retrieve top k numbers while(k>0&&!pq.isEmpty()){ k--; aList.add(pq.poll().getKey()); } return aList; } }