692. Top K Frequent Words

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692. Top K Frequent Words

Description

Difficulty: Medium

Related Topics: Hash Table, String, Trie, Sorting, Heap (Priority Queue), Bucket Sort, Counting

Given an array of strings words and an integer k, return the k most frequent strings.

Return the answer sorted by the frequency from highest to lowest. Sort the words with the same frequency by their lexicographical order.

Example 1:

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Input: words = ["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:

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Input: words = ["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.

Constraints:

  • 1 <= words.length <= 500
  • 1 <= words[i].length <= 10
  • words[i] consists of lowercase English letters.
  • k is in the range [1, The number of **unique** words[i]]

Follow-up: Could you solve it in O(n log(k)) time and O(n) extra space?

Hints/Notes

  • priority queue

Solution

Language: C++

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class Solution {
public:
    vector<string> topKFrequent(vector<string>& words, int k) {
        map<string, int> m;
        for (string word : words) {
            m[word]++;
        }
        auto cmp = [](pair<int, string> lhs, pair<int, string> rhs) {
            if (lhs.first != rhs.first) {
                return lhs.first < rhs.first;
            }
            return lhs.second > rhs.second;
        };
        priority_queue<pair<int, string>, vector<pair<int, string>>, decltype(cmp)> pq(cmp);
        for (auto it : m) {
            string word = it.first;
            int freq = it.second;
            pq.push({freq, word});
        }
        vector<string> res;
        while (k > 0 && !pq.empty()) {
            auto pair = pq.top();
            pq.pop();
            res.push_back(pair.second);
            k--;
        }
        return res;
    }
};