题目如下:

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.

解题思路:本题比较简单,先求出words中每个单词的最小字母的出现频次,并保存到list中。接下来计算queries中每个单词的最小字母的出现频次,并与words中的频次比较。比较的方法可以用二分查找,这样很快就能得到结果。

代码如下:

class Solution(object):
def numSmallerByFrequency(self, queries, words):
"""
:type queries: List[str]
:type words: List[str]
:rtype: List[int]
"""
def calc(word):
min_v = word[0]
dic = {}
for i in word:
dic[i] = dic.setdefault(i,0) + 1
min_v = min(min_v,i)
return dic[min_v]
words_count = []
for word in words:
words_count.append(calc(word)) words_count.sort() res = []
import bisect
for query in queries:
count = calc(query)
inx = bisect.bisect_right(words_count,count)
res.append(len(words_count) - inx)
return res

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