## 1641. 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`

## Rust Solution

``````struct Solution;

impl Solution {
fn count_vowel_strings(n: i32) -> i32 {
let n = n as usize;
let mut res = 0;
for i in 0..5 {
res += Self::dp(i, n);
}
res
}

fn dp(i: i32, n: usize) -> i32 {
if n == 1 {
1
} else {
let mut res = 0;
for j in 0..=i {
res += Self::dp(j, n - 1);
}
res
}
}
}

#[test]
fn test() {
let n = 1;
let res = 5;
assert_eq!(Solution::count_vowel_strings(n), res);
let n = 2;
let res = 15;
assert_eq!(Solution::count_vowel_strings(n), res);
let n = 33;
let res = 66045;
assert_eq!(Solution::count_vowel_strings(n), res);
}
``````

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