1170. Compare Strings by Frequency of the Smallest Character*-CFANZ编程社区

1170. Compare Strings by Frequency of the Smallest Character*

https://leetcode.com/problems/compare-strings-by-frequency-of-the-smallest-character/

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 <= 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.

C++ 实现 1

首先实现 f(w) 统计最小字母的频率. 之后, 统计 words 中每个字符串的 f(w), 保存到 records, 然后对 records 排序. 对 queries 中的每个字符串, 计算每个字符串的 target = f(q), 查找 records 中 target 的 upper_bound.

class Solution {
private:
    int f(string &w) {
        std::sort(w.begin(), w.end());
        int i = 0;
        while (i < w.size() && w[i] == w[0])
            ++ i;
        return i;
    }
public:
    vector<int> numSmallerByFrequency(vector<string>& queries, vector<string>& words) {
        vector<int> records;
        for (auto &w : words) records.push_back(f(w));
        std::sort(records.begin(), records.end());
        vector<int> res;
        for (auto &q : queries) {
            auto target = f(q);
            int l = 0, h = records.size() - 1;
            while (l <= h) {
                int mid = l + (h - l) / 2;
                if (records[mid] <= target) l = mid + 1;
                else h = mid - 1;
            }
            res.push_back(l >= records.size() ? 0 : records.size() - l);
        }
        return res;
    }
};

C++ 实现 2

upper_bound 可以直接用库函数. 另外 f(w) 也可以使用数组实现. 来自 LeetCode 的 Submission.

class Solution {
public:
     int func(string str) {
        int arr[26]=  {0};
        int s=str.size();
        int temp=0;
        for(int i=0;i<s;i++)
           arr[str[i]-'a']++;
        for(int i=0;i<26;i++)
            if(arr[i]!=0)
            {temp=arr[i];break;}
        return temp;
     }
    
    vector<int> numSmallerByFrequency(vector<string>& queries, vector<string>& words) {
       int len=words.size();
        int sze=queries.size();
        vector<int> res;
        vector<int> vec;
        for(int i=0;i<len;i++)
            vec.push_back(func(words[i]));
        sort(vec.begin(),vec.end());
        for(int i=0;i<sze;i++)
        {
            int val=func(queries[i]);
            res.push_back(vec.end()-upper_bound(vec.begin(),vec.end(),val));
        }
        return res;
    }
};