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:

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(n log k) time and O(n) extra space.

Approach #1: C++.[unordered_map]

class Solution {
public:
    vector<string> topKFrequent(vector<string>& words, int k) {
        if (words.empty()) return {};
        unordered_map<string, int> m;
        for (string word : words) {
            m[word]++;
        }
        vector<pair<string, int>> temp(m.begin(), m.end());
        sort(temp.begin(), temp.end(), cmp);
        vector<string> ans;
        for (int i = 0; i < k; ++i) {
            ans.push_back(temp[i].first);
        }
        return ans;
    }
    
private:
    static bool cmp(pair<string, int> a, pair<string, int> b) {
        if (a.second == b.second) return a.first < b.first;
        return a.second > b.second;
    }
};

　　

Approach #2: Java. [heap]

class Solution {
    public List<String> topKFrequent(String[] words, int k) {
        Map<String, Integer> count = new HashMap();
        for (String word : words) {
            count.put(word, count.getOrDefault(word, 0) + 1);
        }
        PriorityQueue<String> heap = new PriorityQueue<String>(
            (w1, w2)->count.get(w1).equals(count.get(w2)) ?
            w2.compareTo(w1) : count.get(w1) - count.get(w2) );
        
        for (String word : count.keySet()) {
            heap.offer(word);
            if (heap.size() > k) heap.poll();
        }
        
        List<String> ans = new ArrayList();
        while (!heap.isEmpty()) ans.add(heap.poll());
        Collections.reverse(ans);
        return ans;
    }
}

　　

Approach #3: Python.

import collections
import heapq
class Solution:
    # Time Complexity = O(n + nlogk)
    # Space Complexity = O(n)
    def topKFrequent(self, words, k):
        count = collections.Counter(words)
        heap = []
        for key, value in count.items():
            heapq.heappush(heap, Word(value, key))
            if len(heap) > k:
                heapq.heappop(heap)
        res = []
        for _ in range(k):
            res.append(heapq.heappop(heap).word)
        return res[::-1]

class Word:
    def __init__(self, freq, word):
        self.freq = freq
        self.word = word

    def __lt__(self, other):
        if self.freq == other.freq:
            return self.word > other.word
        return self.freq < other.freq

    def __eq__(self, other):
        return self.freq == other.freq and self.word == other.word