23. 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

Solution: (Using Backtracking to generate all possible strings and counting length)

Solution: (DP)

Approach: Count the no of ways from 'u' in each case and check the pattern.

class Solution
{
public:
    int countVowelStrings(int n)
    {
        
        //No of ways when n = 1 
        vector<int> dp{5, 4, 3, 2, 1, 0};
        
        // for higher n
        // the number of ways(i) = number of ways(i) + number of ways(i+1)
        
        for (int i = 1; i < n; i++)
        {
            for (int j = 4; j >= 0; j--)
            {
                dp[j] += dp[j + 1];
            }
        }

        return dp[0];
    }
};