# 1170. Compare Strings by Frequency of the Smallest Character 比较字符串最小字母出现频次

@TOC

## # 题目描述

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. `1 <= queries.length <= 2000`
2. `1 <= words.length <= 2000`
3. `1 <= queries[i].length, words[i].length <= 10`
4. `queries[i][j], words[i][j]` are English lowercase letters.

## # 解题方法

### # 双重循环

C++代码如下：

``````class Solution {
public:
vector<int> numSmallerByFrequency(vector<string>& queries, vector<string>& words) {
vector<int> qs, ws;
for (string& q : queries) {
qs.push_back(getFrequency(q));
}
for (string& w : words) {
ws.push_back(getFrequency(w));
}
vector<int> res;
for (int q : qs) {
int count = 0;
for (int w : ws) {
if (w > q)
count++;
}
res.push_back(count);
}
return res;
}
int getFrequency(string& word) {
vector<int> counts(26, 0);
for (char w : word) {
counts[w - 'a']++;
}
for (int i = 0; i < 26; ++i) {
if (counts[i] != 0)
return counts[i];
}
return 0;
}
};
``````

## # 日期

2019 年 8 月 31 日 —— 赶在月底做个题