[LeetCode] 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.

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.

比较字符串最小字母出现频次。

我们来定义一个函数 f(s)，其中传入参数 s 是一个非空字符串；该函数的功能是统计 s  中（按字典序比较）最小字母的出现频次。

例如，若 s = "dcce"，那么 f(s) = 2，因为最小的字母是 "c"，它出现了 2 次。

现在，给你两个字符串数组待查表 queries 和词汇表 words，请你返回一个整数数组 answer 作为答案，其中每个 answer[i] 是满足 f(queries[i]) < f(W) 的词的数目，W 是词汇表 words 中的词。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/compare-strings-by-frequency-of-the-smallest-character
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这道题暴力解不难想，但是暴力解是O(n^2)级别的，很显然会超时。我这里提供一个二分法的思路。这道题其实不是很难但是题意写的不是很清楚，f函数 其实需要输出的是字典序最小的出现次数不为0的字母的出现次数。

首先根据题意我们去实现 f函数 ，这个不是很难。我们创建两个临时数组 int[] q 和 int[] w ，分别把queries和words里面的每个单词跑过F函数的结果存储起来，同时我们把 int[] w 排序，这样找最后结果的时候可以用到二分法。

时间O(nlogn)

空间O(n)

Java实现

class Solution {
    public int[] numSmallerByFrequency(String[] queries, String[] words) {
        int[] q = new int[queries.length];
        int[] w = new int[words.length];
        for (int i = 0; i < q.length; i++) {
            q[i] = helper(queries[i]);
        }
        for (int j = 0; j < w.length; j++) {
            w[j] = helper(words[j]);
        }
        Arrays.sort(w);
        int[] res = new int[q.length];
        for (int i = 0; i < q.length; i++) {
            int left = 0;
            int right = w.length - 1;
            while (left <= right) {
                int mid = left + (right - left) / 2;
                if (q[i] >= w[mid]) {
                    left = mid + 1;
                } else {
                    right = mid - 1;
                }
            }
            res[i] = w.length - left;
        }
        return res;
    }

    private int helper(String s) {
        int[] count = new int[26];
        for (int i = 0; i < s.length(); i++) {
            count[s.charAt(i) - 'a']++;
        }

        for (int i = 0; i < 26; i++) {
            if (count[i] != 0) {
                return count[i];
            }
        }
        return 0;
    }
}

LeetCode 题目总结