[LC] 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.

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.

 1 import collections
 2 import heapq
 3 
 4 class Element(object):
 5     def __init__(self, word, freq):
 6         self.word = word
 7         self.freq = freq
 8     
 9     def __lt__(self, other):
10         if self.freq != other.freq:
11             return self.freq < other.freq
12         return other.word < self.word
13 
14 class Solution(object):
15     def topKFrequent(self, words, k):
16         """
17         :type words: List[str]
18         :type k: int
19         :rtype: List[str]
20         """
21         my_dict = {}
22         for word in words:
23             if word in my_dict:
24                 my_dict[word] += 1
25             else:
26                 my_dict[word] = 1 
27         
28         freqs = []
29         for word, count in my_dict.items():
30             heapq.heappush(freqs, (Element(word, count)))
31             if len(freqs) > k:
32                 heapq.heappop(freqs)
33         res = []
34         for _ in range(k):
35             res.append(heapq.heappop(freqs).word)
36         res.reverse()
37         return res
38     
39

Solution 2:

O(klogN)

class Solution {
    public List<String> topKFrequent(String[] words, int k) {
    Map<String, Integer> map = new HashMap<>();
    for (String s: words) {
      map.put(s, map.getOrDefault(s, 0) + 1);
    }
    // keep a top frequency heap
    PriorityQueue<Map.Entry<String, Integer>> pq = new PriorityQueue<>((a, b) -> 
      a.getValue() == b.getValue() ? a.getKey().compareTo(b.getKey()): b.getValue() - a.getValue()
    );
    pq.addAll(map.entrySet());
    List<String> res = new ArrayList<>();
    int i = 0;
    while (i < k) {
        res.add(pq.poll().getKey());
        i += 1;
    }
    return res;        
    }
}