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.

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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").

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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.

這速度真夠慢的。。。。

先寫一個函數(shù)判斷是否小于，一開始是用26長度的整數(shù)數(shù)組，報錯了一次，于是直接把string改為char array了，sort一次，然后遍歷去比對。

另個函數(shù)就是直接去遍歷數(shù)據，放到數(shù)組當中，再返回即可。

Runtime1675 ms

Beats

5.11%

Memory44 MB

Beats

65.53%