. [Swift] LeetCode1170 compare strings smallest letter appears frequency | Compare Strings by Frequency of the Smallest Character

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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.

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We define a function  f(s), passing in parameters  s is a non-empty string; function is the function of statistics  s  appear (by comparing lexicographically) smallest letter frequency.

For example, if  s = "dcce", then  f(s) = 2, because the smallest letters  "c", it appears twice.

Now, to give you two string arrays to be look-up table  queries and glossary  words, please return an array of integers  answer as the answer, each of which  answer[i] is to meet  f(queries[i]) <  f(W) the number of words, W it is a glossary  words of words.

 

Example 1:

Input: queries = [ "cbd"] , words = [ "zaaaz"] 
Output: [1] 
Explanation: Query f ( "cbd") = 1 , and f ( "zaaaz") = 3 so that f ( "cbd") <f ( "zaaaz").

Example 2:

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

 

prompt:

• 1 <= queries.length <= 2000
• 1 <= words.length <= 2000
• 1 <= queries[i].length, words[i].length <= 10
• queries[i][j],  words[i][j] All lowercase letters