Top k Frequent Words

Problem

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.

Requirements

  • return the k most frequent elements
  • 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
  • words contain only lowercase letters
  • solve it in O(n log k) time and O(n) extra space

Approach

We can use a map for gathering the counts of all the words. This can be done in O(n) time and will use O(n) space. Once we have this, we can use a priority queue with a custom comparator.

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class Solution {

public:
  vector<string> topKFrequent(vector<string>& words, int k) {
    using word_map_t = unordered_map<string, int>;
    using pair_t = pair<string, int>;
    using vec_t = vector<pair_t>;

    word_map_t word_map;
    for (auto w : words) {
      ++word_map[w];
    }

    auto cmp = [&](pair_t& a, pair_t& b) {
      return (a.second < b.second || (a.second == b.second && a.first > b.first));
    };

    priority_queue<pair_t, vec_t, decltype(cmp)> pq(cmp);

    for (auto w : word_map) {
      pq.emplace(w.first, w.second);
    }
    vector<string> res;
    while (k > 0 && !pq.empty()) {
      //res.insert(res.begin(), pq.top().first);
      res.push_back(pq.top().first);
      pq.pop();
      k--;
    }

    return res;
  }
};