692. Top K Frequent Words

692. Top K Frequent Words

Given a non-empty list of words, return the k most frequent elements.

Your answer should be sorted by frequency from highest to lowest. If two words have
the same frequency, then the word with the lower alphabetical order comes first.

Example 1:

Input: ["i", "love", "leetcode", "i", "love", "coding"], k = 2
Output: ["i", "love"]
Explanation: "i" and "love" are the two most frequent words.
Note that "i" comes before "love" due to a lower alphabetical order.

• Sorting

public List<String> topKFrequentSort(String[] words, int k) {
    HashMap<String, Integer> map = new HashMap<>();
    for(String s : words) {
        if(!map.containsKey(s)) map.put(s, 0);
        map.put(s, map.get(s)+1);
    }
    ArrayList<Map.Entry<String,Integer>> arr = new ArrayList<>(map.entrySet());
    Collections.sort(arr, (x,y)->{
        int r = Integer.compare(y.getValue(), x.getValue());
        if(r == 0) return x.getKey().compareTo(y.getKey());
        return r;
    });
    ArrayList<String> result = new ArrayList<>();

    for(int i = 0; i < k; i++) result.add(arr.get(i).getKey());
    return result;
}

• PriorityQueue

  public List<String> topKFrequent(String[] words, int k) {
    HashMap<String, Integer> map = new HashMap<>();
    for(String s : words) {
        map.put(s, map.getOrDefault(s, 0)+1);
    }
    PriorityQueue<Map.Entry<String, Integer>> q = new PriorityQueue<>((x,y)->{
        int r = Integer.compare(x.getValue(), y.getValue());
        if(r == 0) return y.getKey().compareTo(x.getKey());
        return r;
    });
  
    for(Map.Entry<String, Integer> e : map.entrySet()) {
        if(q.size() < k) q.offer(e);
        else if(q.peek().getValue() < e.getValue() || q.peek().getValue() == e.getValue() && q.peek().getKey().compareTo(e.getKey()) > 0) {
            q.poll();
            q.offer(e);
        }
    }
    LinkedList<String> result = new LinkedList<>();
    while(!q.isEmpty()) {
        result.addFirst(q.poll().getKey());
    }
    return result;
  }