1170. Compare Strings by Frequency of the Smallest Character

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.

• 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.

o（min(n, m)）首先肯定是要统计每个word的f值，由于每个单词的长度不超过10，那么f值肯定也不超过10，那就维护一个cnt表示所有word的f值出现的次数，再对这个cnt从后往前求一个前缀和，得到cnt[i]表示大于等于i的f值出现的次数

class Solution(object):
    def numSmallerByFrequency(self, queries, words):
        """
        :type queries: List[str]
        :type words: List[str]
        :rtype: List[int]
        """
        def f(word):
            cnt = [0] * 26
            for c in word:
                cnt[ord(c) - ord('a')] += 1
            for i in range(26):
                if cnt[i] != 0:
                    return cnt[i]
                
        cnt = [0] * 12
        for word in words:
            cnt[f(word)] += 1
        for i in range(10, 0, -1):
            cnt[i] += cnt[i + 1]
        ans = []
        for query in queries:
            ans.append(cnt[f(query) + 1])
        return ans