README.md

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

Solutions (Ruby)

1. Dynamic Programming

# @param {Integer} n
# @return {Integer}
def count_vowel_strings(n)
  dp = [1] * 5

  (1...n).each do |_|
    dp[1] += dp[0]
    dp[2] += dp[1]
    dp[3] += dp[2]
    dp[4] += dp[3]
  end

  dp.sum
end

Solutions (Rust)

1. Dynamic Programming

impl Solution {
    pub fn count_vowel_strings(n: i32) -> i32 {
        let mut dp = [1; 5];

        for _ in 1..n {
            dp[1] += dp[0];
            dp[2] += dp[1];
            dp[3] += dp[2];
            dp[4] += dp[3];
        }

        dp.iter().sum()
    }
}