# 1641. Count Sorted Vowel Strings

#### Medium

***

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`&#x20;

```python
class Solution:
    def countVowelStrings(self, n: int) -> int:
        dp = [[1]*6] + [[0]*(6) for _ in range(n)]
        for i in range(1, n+1):
            for k in range(1,6):
                dp[i][k] = dp[i-1][k] + dp[i][k-1]
        return dp[n][k]
    
    # TLE
    def countVowelStringsOld(self, n: int) -> int:
        vowels = ['a','e','i', 'o', 'u']
        def recursion(vowels, start, path, result):
            if len(path) == n:
                result.append(path[:])
                return
            for index in range(start, len(vowels)):
                path.append(vowels[index])
                # No index+1 since vowels can repeat
                recursion(vowels, index, path, result)
                path.pop()
        result = []
        recursion(vowels, 0, [], result)
        return len(result)
```