題目

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.

題目大意

我們來定義一個函數 f(s)，其中傳入參數 s 是一個非空字符串；該函數的功能是統計 s  中（按字典序比較）最小字母的出現頻次。

例如，若 s = "dcce"，那麼 f(s) = 2，因為最小的字母是 "c"，它出現了 2 次。

現在，給你兩個字符串數組待查表 queries 和詞彙表 words，請你返回一個整數數組 answer 作為答案，其中每個 answer[i] 是滿足 f(queries[i]) < f(W) 的詞的數目，W 是詞彙表 words 中的詞。

提示：

• 1 <= queries.length <= 2000
• 1 <= words.length <= 2000
• 1 <= queries[i].length, words[i].length <= 10
• queries[i][j], words[i][j] 都是小寫英文字母

解題思路

• 給出 2 個數組，queries 和 words，針對每一個 queries[i] 統計在 words[j] 中滿足 f(queries[i]) < f(words[j]) 條件的 words[j] 的個數。f(string) 的定義是 string 中字典序最小的字母的頻次。
• 先依照題意，構造出 f() 函數，算出每個 words[j] 的 f() 值，然後排序。再依次計算 queries[i] 的 f() 值。針對每個 f() 值，在 words[j] 的 f() 值中二分搜索，查找比它大的值的下標 k，n-k 即是比 queries[i] 的 f() 值大的元素個數。依次輸出到結果數組中即可。 �次輸出到結果數組中即可。

參考代碼

package leetcode

import "sort"

func numSmallerByFrequency(queries []string, words []string) []int {
	ws, res := make([]int, len(words)), make([]int, len(queries))
	for i, w := range words {
		ws[i] = countFunc(w)
	}
	sort.Ints(ws)
	for i, q := range queries {
		fq := countFunc(q)
		res[i] = len(words) - sort.Search(len(words), func(i int) bool { return fq < ws[i] })
	}
	return res
}

func countFunc(s string) int {
	count, i := [26]int{}, 0
	for _, b := range s {
		count[b-'a']++
	}
	for count[i] == 0 {
		i++
	}
	return count[i]
}