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].

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.

public class Solution {
    public List<Integer> topKFrequent(int[] nums, int k) {
        //bucket: an array of list  
        List<Integer>[] bucket = new List[nums.length+1];
        Map<Integer, Integer> frequencyMap = new HashMap<Integer, Integer>(); 
        for (int n: nums) {
            //getOrDefaul: Returns the value to which the specified key 
            //is mapped, or defaultValue (0) if this map contains no 
            //mapping for the key.
            frequencyMap.put(n, frequencyMap.getOrDefault(n,0)+1); 
        for (int key:frequencyMap.keySet()) {
            int frequency = frequencyMap.get(key);
            //high index stands for high frequency 
            if (bucket[frequency] == null) {
                bucket[frequency] = new ArrayList<>(); 
        List<Integer> res = new ArrayList<>(); 
        for (int pos = bucket.length - 1; pos >= 0 && res.size() < k; pos--) {
            //start with the largest index: 
            if (bucket[pos] != null) {
        return res;

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