## DEV Community

Abhishek Chaudhary

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# 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]
``````