## [692\. Top K Frequent Words](https://leetcode.com/problems/top-k-frequent-words/)

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:**

**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:**

**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?

class Solution {
public:
    vector<string> topKFrequent(vector<string>& words, int k) {
        // res 最多就 k 的長度
        vector<string> res(k);
        
        // 計算每個單字的頻率
        unordered_map<string, int> freq;
        for (auto word : words) ++freq[word];
        
        auto cmp = [](pair<string, int>& a, pair<string, int>& b) {
            // a.second > b.second 比較頻率
            // priority_queue 和 sort custom compare 的順序相反
            // 如果頻率是由小排到大的話，會是 a.second > b.second

            // 如果出現頻率一樣的話，a.second == b.second && a.first < b.first
            // 比較字串
            return a.second > b.second || (a.second == b.second && a.first < b.first);
        };
        priority_queue<pair<string, int>, vector<pair<string, int>>, decltype(cmp)> pq(cmp);
        
        for (auto f : freq) {
            pq.push(f);
            // 維持 k 的長度，如果太長就 pop 掉，從頻率少的開始 popup
            if (pq.size() > k) pq.pop();
        }
        
        // 因為這時候的 priority queue 是從小排到大
        // 所以從 res 的最後面開始裝
        for (int i = res.size() - 1; i >= 0; --i) {
            res[i] = pq.top().first; pq.pop();
        }
        return res;
    }
};

:::success
- 時間複雜度：$O(N)$
- 空間複雜度：$O(N)$
:::