# 1641. Count Sorted Vowel Strings

[Link](https://leetcode.com/problems/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

```java
    public int countVowelStrings(int n) {
        int[][] dp = new int[n+1][6];
        for(int i = 1; i <= n; i++){
            for(int k = 1; k < 6; k++){
                //dp[i][k-1], length i and not include vowels[k]
                dp[i][k] = dp[i][k-1] + ( i > 1 ? dp[i-1][k] : 1); 
                                                  //dp[i-1][k] : combination count of s[0....i], 
                                                  //where we fixed the s[i] is the lacking vowels[k]    
                                                  // so dp[i-1][k] will be the count of including voewls[k] 
            }
            
        }
        return dp[n][5];
    }
```
