692 Top K Frequent Words

692. Top K Frequent Words

1. Question

Given a non-empty list of words, return thekmost 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:

1. You may assume k is always valid, 1 ≤ k ≤ number of unique elements.

2. Input words contain only lowercase letters.

Follow up:

1. Try to solve it in O(nlogk) time and O(n) extra space.

2. Implementation

(1) Trie + PriorityQueue

class Solution {
    public List<String> topKFrequent(String[] words, int k) {
        List<String> res = new ArrayList<>();

        Map<String, Integer> map = new HashMap<>();

        for (String word : words) {
            map.put(word, map.getOrDefault(word, 0) + 1);
        }

        PriorityQueue<WordInfo> maxHeap = new PriorityQueue<>();

        for (String key : map.keySet()) {
            maxHeap.add(new WordInfo(map.get(key), key));
        }

        while (k-- > 0 && !maxHeap.isEmpty()) {
            res.add(maxHeap.remove().word);
        }

        return res;
    }

    class WordInfo implements Comparable<WordInfo> {
        int freq;
        String word;

        public WordInfo(int freq, String word) {
            this.freq = freq;
            this.word = word;
        }

        public int compareTo(WordInfo that) {
            return this.freq == that.freq ? word.compareTo(that.word) : that.freq - this.freq;
        }
    }
}

3. Time & Space Complexity

Trie + PriorityQueue: 时间复杂度O(nlogk), n是word的个数，空间复杂度O(n)