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 collections import Counter
class Solution(object):
    def numSmallerByFrequency(self, queries, words):
        """
        :type queries: List[str]
        :type words: List[str]
        :rtype: List[int]
        """
        qs,ws = [],[]
        for q in queries: qs.append(Counter(q)[min(q)])
        for w in words: ws.append(Counter(w)[min(w)])
        d=Counter(ws)
        helper = [0]*2005
        for i in range(2003, -1, -1):
            helper[i]=helper[i+1]
            if i in d: helper[i]+=d[i]
        
        res=[]
        for i in qs:
            res.append(helper[i]-d[i])
        return res