# 0692. Top K Frequent Words

<https://leetcode.com/problems/top-k-frequent-words>

## Description

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] <= 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?

## ac

```java
class Solution {
    public List<String> topKFrequent(String[] words, int k) {
        List<String> res = new ArrayList<>();
        // edge cases
        if (words == null || words.length == 0 || k < 1) return res;

        Map<String, Integer> map = new HashMap<>();
        for (String word : words) {
            map.put(word, map.getOrDefault(word, 0)+1);
        }

        PriorityQueue<Map.Entry<String,Integer>> pq = new PriorityQueue<>((a, b) -> {
            return a.getValue() == b.getValue() ? a.getKey().compareTo(b.getKey()) : b.getValue() - a.getValue();
        });

        for (Map.Entry<String, Integer> e : map.entrySet()) {
            pq.offer(e);
        }

        for (int i = 0; i < k; i++) {
            res.add(pq.poll().getKey());
        }

        return res;
    }
}

// same as https://leetcode.com/problems/top-k-frequent-elements
```