1170. Compare Strings by Frequency of the Smallest Character

https://leetcode.com/problems/compare-strings-by-frequency-of-the-smallest-character

Description

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.

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