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# @lc app=leetcode.cn id=692 lang=python
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# [692] 前K个高频单词
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# https://leetcode-cn.com/problems/top-k-frequent-words/description/
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# algorithms
# Medium (39.56%)
# Likes:    37
# Dislikes: 0
# Total Accepted:    2K
# Total Submissions: 5.1K
# Testcase Example:  '["i", "love", "leetcode", "i", "love", "coding"]\n2'
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# 给一非空的单词列表，返回前 k 个出现次数最多的单词。
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# 返回的答案应该按单词出现频率由高到低排序。如果不同的单词有相同出现频率，按字母顺序排序。
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# 示例 1：
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# 输入: ["i", "love", "leetcode", "i", "love", "coding"], k = 2
# 输出: ["i", "love"]
# 解析: "i" 和 "love" 为出现次数最多的两个单词，均为2次。
# ⁠   注意，按字母顺序 "i" 在 "love" 之前。
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# 示例 2：
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# 输入: ["the", "day", "is", "sunny", "the", "the", "the", "sunny", "is", "is"],
# k = 4
# 输出: ["the", "is", "sunny", "day"]
# 解析: "the", "is", "sunny" 和 "day" 是出现次数最多的四个单词，
# ⁠   出现次数依次为 4, 3, 2 和 1 次。
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# 注意：
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# 假定 k 总为有效值， 1 ≤ k ≤ 集合元素数。
# 输入的单词均由小写字母组成。
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# 扩展练习：
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# 尝试以 O(n log k) 时间复杂度和 O(n) 空间复杂度解决。
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class Solution(object):
    def topKFrequent(self, words, k):
        """
        :type words: List[str]
        :type k: int
        :rtype: List[str]
        """
        map = {}
        for i in words:
            map[i] = map.get(i, 0) + 1
        ret = sorted(map, key=lambda word: (-map[word], word))
        return ret[:k]


# print(Solution().topKFrequent(
#     ["the", "day", "is", "sunny", "the", "the", "the", "sunny", "is", "is"],
#     2))
