Count Sorted Vowel Strings

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 

SOLUTION:

class Solution:
    def cvow(self, n):
        if n == 1:
            return [5, 4, 3, 2, 1]
        curr = self.cvow(n - 1)
        op = [curr[-1]]
        for i in range(3, -1, -1):
            op.append(op[-1] + curr[i])
        return op[::-1]

    def countVowelStrings(self, n: int) -> int:
        return self.cvow(n)[0]