692. Top K Frequent Words

Question

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.
Example 2:
Input: ["the", "day", "is", "sunny", "the", "the", "the", "sunny", "is", "is"], k = 4
Output: ["the", "is", "sunny", "day"]
Explanation: "the", "is", "sunny" and "day" are the four most frequent words,
    with the number of occurrence being 4, 3, 2 and 1 respectively.

Note:

You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
Input words contain only lowercase letters. Follow up:
Try to solve it in O(n log k) time and O(n) extra space.

Solution

Method 1: TreeMap + PriorityQueue

class Solution {
    private static class Pair{
        String s;
        int count;
        public Pair(String s, int count){
            this.s = s;
            this.count = count;
        }
    }
    public List<String> topKFrequent(String[] words, int k) {
        TreeMap<String, Integer> treeMap = new TreeMap<>(new Comparator<String>(){  //time complexity: O(nlogN)
            @Override
            public int compare(String a, String b){
                return a.compareTo(b);
            }
        });
        for(String word : words){
            treeMap.put(word, treeMap.getOrDefault(word, 0) + 1);
        }

        PriorityQueue<Pair> pq = new PriorityQueue<>(new Comparator<Pair>(){    //time complexity: O(nlogN)
            @Override
            public int compare(Pair p1, Pair p2){
                return p2.count == p1.count ? p1.s.compareTo(p2.s) : p2.count - p1.count;
            }
        });
        for(Map.Entry<String, Integer> entry : treeMap.entrySet()){
            pq.offer(new Pair(entry.getKey(), entry.getValue()));
        }
        List<String> result = new ArrayList<>();
        for(int i = 0; i < k; i++){
            result.add(pq.poll().s);
        }        
        return result;
    }
}