LeetCode 1170.Compare Strings by

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.

Solution:

class Solution:
    def numSmallerByFrequency(self, queries: List[str], words: List[str]) -> List[int]:
        word_freq = sorted([w.count(min(w)) for w in words])
        res = []
        for query in queries:
            q_count = query.count(min(query))
            res.append(len(words) - bisect.bisect(word_freq, q_count))
        return res

Explanation:
Do a count of min word in the words array
Get the minimum letter
Count frequency of letter
Do a binary search and find the items that have more frequency of min char