[leetcode] 1170. Compare Strings by Frequency of the Smallest Character

[leetcode] 1170. Compare Strings by Frequency of the Smallest Character

Description

Let’s define a function f(s) over a non-empty string s, which calculates the frequency of the smallest character in s. For example, if s = “dcce” then f(s) = 2 because the smallest character is “c” and its frequency is 2.

Now, given string arrays queries and words, return an integer array answer, where each answer[i] is the number of words such that f(queries[i]) < f(W), where W is a word in words.

Example 1:

Input: queries = ["cbd"], words = ["zaaaz"]Output: [1]Explanation: On the first query we have f("cbd") = 1, f("zaaaz") = 3 so f("cbd") < f("zaaaz").

Example 2:

Input: queries = ["bbb","cc"], words = ["a","aa","aaa","aaaa"]Output: [1,2]Explanation: On the first query only f("bbb") < f("aaaa"). On the second query both f("aaa") and f("aaaa") are both > f("cc").

Constraints:

1 <= queries.length <= 2000.1 <= words.length <= 2000.1 <= queries[i].length, words[i].length <= 10.queries[i][j], words[i][j] are English lowercase letters.

分析

题目的意思是：给定一个query数组和一个words，统计query和words里面最小字符的频率，然后query中的每一个字符不超过words里面最小字符的频率的数量。这道题一开始要统计词频，query和words都要统计，之后就是一个比较了，用sum函数能够省很多事。

代码

class Solution: def numSmallerByFrequency(self, queries: List[str], words: List[str]) -> List[int]: q=[] for query in queries: cnt=0 prev='' for i in range(len(query)): if(i==0): prev=query[i] cnt=1 elif(query[i] q1 for j in w) res.append(count) return res

参考文献

​​[LeetCode] easy python solution for beginners : )​​)

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