LeetCode-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

Analysis:

We can use bucket sort style solution: Since the largest frequency is N, we can use a size N array to store the element with each possible frequency.

Solution:

public class Solution {
    public List<Integer> topKFrequent(int[] nums, int k) {
        List<Integer> res = new ArrayList<Integer>();
        if (nums.length==0 || k==0) return res;
        
        List<Integer>[] freqList = new List[nums.length+1];
        Map<Integer,Integer> freqMap = new HashMap<Integer,Integer>();
        
        // Get frequencey map
        for (int num: nums){
            freqMap.put(num,freqMap.getOrDefault(num,0)+1);
        }
        
        // According to each num's frequency, put it into frequency list
        for (int num: freqMap.keySet()){
            int freq = freqMap.get(num);
            if (freqList[freq]==null){
                freqList[freq] = new ArrayList<Integer>();
            }
            freqList[freq].add(num);
        }
        
        int left = k;
        for (int i=freqList.length-1;i>=0 && left>0;i--){
            if (freqList[i]!=null){
                if (left>=freqList[i].size()){
                    res.addAll(freqList[i]);
                    left -= freqList[i].size();
                } else {
                    for (int j=0;j<left;j++) res.add(freqList[i].get(j));
                    break;
                }
            }
        }
        
        return res;
    }
}