我们来定义一个函数 f(s),其中传入参数 s 是一个非空字符串;该函数的功能是统计 s 中(按字典序比较)最小字母的出现频次。
例如,若 s = "dcce",那么 f(s) = 2,因为最小的字母是 "c",它出现了 2 次。
现在,给你两个字符串数组待查表 queries 和词汇表 words,请你返回一个整数数组 answer 作为答案,其中每个 answer[i] 是满足 f(queries[i]) < f(W) 的词的数目,W 是词汇表 words 中的词。
示例 1:
输入:queries = ["cbd"], words = ["zaaaz"]
输出:[1]
解释:查询 f("cbd") = 1,而 f("zaaaz") = 3 所以 f("cbd") < f("zaaaz")。
示例 2:
输入:queries = ["bbb","cc"], words = ["a","aa","aaa","aaaa"]
输出:[1,2]
解释:第一个查询 f("bbb") < f("aaaa"),第二个查询 f("aaa") 和 f("aaaa") 都 > f("cc")。
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/compare-strings-by-frequency-of-the-smallest-character
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# leetcode 1170
# 通俗易懂,平淡无奇的解法
from typing import List
from collections import Counter
class Solution:
def numSmallerByFrequency(self, queries: List[str], words: List[str]) -> List[int]:
list2 = []
list3 = []
for i in queries:
list2.append(self.countNum(i))
for j in words:
list3.append(self.countNum(j))
res = []
for i in range(len(list2)):
count = 0
for j in range(len(list3)):
if list2[i] < list3[j]:
count += 1
res.append(count)
return res
def countNum(self, strs):
temp = "".join(sorted(strs))
dict2 = dict(Counter(temp))
res = 0
for k, v in dict2.items():
if k == temp[0]:
res = v
return res
if __name__ == "__main__":
s = Solution()
queries = ["bbb", "cc"]
words = ["a", "aa", "aaa", "aaaa"]
print(s.numSmallerByFrequency(queries, words))