LeetCode解题报告（270）-- 1641. Count Sorted Vowel Strings

Problem

Given an integer n, return the number of strings of length n that consist only of vowels (a, e, i, o, u) and are lexicographically sorted.

A string s is lexicographically sorted if for all valid i, s[i] is the same as or comes before s[i+1] in the alphabet.

Example 1:

Input: n = 1
Output: 5
Explanation: The 5 sorted strings that consist of vowels only are ["a","e","i","o","u"].

Example 2:

Input: n = 2
Output: 15
Explanation: The 15 sorted strings that consist of vowels only are
["aa","ae","ai","ao","au","ee","ei","eo","eu","ii","io","iu","oo","ou","uu"].
Note that "ea" is not a valid string since 'e' comes after 'a' in the alphabet.

Example 3:

Input: n = 33
Output: 66045

Constraints:

• 1 <= n <= 50

---

Analysis

  题目给出一个数字n，代表字符串的长度，规定字符串只能由元音字母构成，而且必须是字典序，问一共有多少种组合。这道题目有暴力求解，也能够使用动态规划，但这里我分享一个利用数学推理来解决的方法。

  对于最终的字符串，我们假假设选取了五个原因字符分别有a、e、i、o、u个。实际上我们可以想象成，有n个小球，我们需要给这些小球标上这些元音字符（5种且按照字典序），问有多少种分割的方法。为了分出5种类型，我们需要插入4块隔板，插入的位置有n - 1个。为什么是n - 1呢？因为这样保证了每种类型都至少有一个，在这个情况下，总数是$C^{m-1}_{n-1}$那么问题来了，题目的条件是可以有部分元音不出现，所以我们希望的是隔板之间可以为空的情况。

  我们如何解决这个问题呢？其实很简单，我们只需要多要m个小球，首先分到m个盒子，然后再按照上面的情况处理，最后再把这m个小球都拿走，得到的结果就是n个球分到m个盒子，盒子可以为空的情况了，所以答案就是$C_{n+m-1}^{m - 1}$。

---

Solution

  无

---

Code

class Solution {
public:
    int countVowelStrings(int n) {
        return ((n + 4) * (n + 3) * (n + 2)  * (n + 1)) / 24;
    }
};

---

Summary

  这道题目从数学的角度求解，思考过程比较复杂，也比较考察排列组合方面的知识，但是coding量就非常少了。这道题这道题目的分享到这里，谢谢您的支持！