Leetcode1170. Compare Strings by Frequency of the Smallest Chara

Let the function f(s) be the frequency of the lexicographically smallest character in a non-empty string s. For example, if s = "dcce" then f(s) = 2 because the lexicographically smallest character is 'c', which has a frequency of 2.

You are given an array of strings words and another array of query strings queries. For each query queries[i], count the number of words in words such that f(queries[i]) < f(W) for each W in words.

Return an integer array answer, where each answer[i] is the answer to the ith query.

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] consist of lowercase English letters.

这速度真够慢的。。。。

先写一个函数判断是否小于，一开始是用26长度的整数数组，报错了一次，于是直接把string改为char array了，sort一次，然后遍历去比对。

另个函数就是直接去遍历数据，放到数组当中，再返回即可。

Runtime1675 ms

Beats

5.11%

Memory44 MB

Beats

65.53%