1170. 比较字符串最小字母出现频次

题目链接	1170. 比较字符串最小字母出现频次
思路	题意不易理解；排序+二分（upper_bound）
题解链接	Python简洁解法
关键点	预先处理words
时间复杂度	\(O((n+m)p)\)
空间复杂度	\(O(1)\)

代码实现：

class Solution:def numSmallerByFrequency(self, queries: List[str], words: List[str]) -> List[int]:nums = sorted(s.count(min(s)) for s in words)n = len(nums)def upper_bound(val):left, right = -1, nwhile left+1<right:mid = (left+right) // 2if nums[mid] > val:right = midelse:left = midreturn rightreturn [n-upper_bound(q.count(min(q))) for q in queries]

Python-官方库

class Solution:def numSmallerByFrequency(self, queries: List[str], words: List[str]) -> List[int]:nums = sorted(s.count(min(s)) for s in words)n = len(nums)return [n-bisect_right(nums, q.count(min(q))) for q in queries]