"""
    Tag: string, array, integer

    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.
"""
from typing import List
from collections import Counter
from bisect import bisect


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


assert Solution().numSmallerByFrequency(["cbd"], ["zaaaz"]) == [1]
assert Solution().numSmallerByFrequency(
    ["bbb", "cc"], ["a", "aa", "aaa", "aaaa"]) == [1, 2]
print('Tests Passed!!')

d7532ba3fc27cf7a2802352b9d0ff37ddae6bc8d
<mask token>


class Solution:
    <mask token>


<mask token>

<mask token>


class Solution:

    def numSmallerByFrequency(self, queries: List[str], words: List[str]
        ) ->List[int]:
        words_freq = sorted([word.count(min(word)) for word in words])
        return [(len(words) - bisect(words_freq, query.count(min(query)))) for
            query in queries]


<mask token>

<mask token>


class Solution:

    def numSmallerByFrequency(self, queries: List[str], words: List[str]
        ) ->List[int]:
        words_freq = sorted([word.count(min(word)) for word in words])
        return [(len(words) - bisect(words_freq, query.count(min(query)))) for
            query in queries]


assert Solution().numSmallerByFrequency(['cbd'], ['zaaaz']) == [1]
assert Solution().numSmallerByFrequency(['bbb', 'cc'], ['a', 'aa', 'aaa',
    'aaaa']) == [1, 2]
print('Tests Passed!!')

<mask token>
from typing import List
from collections import Counter
from bisect import bisect


class Solution:

    def numSmallerByFrequency(self, queries: List[str], words: List[str]
        ) ->List[int]:
        words_freq = sorted([word.count(min(word)) for word in words])
        return [(len(words) - bisect(words_freq, query.count(min(query)))) for
            query in queries]


assert Solution().numSmallerByFrequency(['cbd'], ['zaaaz']) == [1]
assert Solution().numSmallerByFrequency(['bbb', 'cc'], ['a', 'aa', 'aaa',
    'aaaa']) == [1, 2]
print('Tests Passed!!')

"""
    Tag: string, array, integer

    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.
"""
from typing import List
from collections import Counter
from bisect import bisect


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


assert Solution().numSmallerByFrequency(["cbd"], ["zaaaz"]) == [1]
assert Solution().numSmallerByFrequency(
    ["bbb", "cc"], ["a", "aa", "aaa", "aaaa"]) == [1, 2]
print('Tests Passed!!')
