Count Sorted Vowel Strings

Problem Statement

Given an integer n, the task is to calculate the number of strings of length n that are made up solely of the vowels: a, e, i, o, u. These strings must adhere to a specific order rule to be considered valid; they must be lexicographically sorted. A string s meets this criterion if each character s[i] is either the same as or comes before s[i+1] in the alphabet. This kind of problem highlights combinatorial logic and the application of combinatorics in string formation under set constraints.

Examples

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

Approach and Intuition

• The examples provided clarify how the outputs are derived based on the input value n.

  • For n = 1, every single vowel is a valid string (i.e.,a, e, i, o, u).
  • For n = 2, combinations of vowels that are in alphabetical order form valid strings such as aa, ae, ai and so on.

• The core observation is that for a string of length n to be lexicographically sorted using only the vowels, we are essentially placing n indistinguishable items (vowels) into 5 distinguishable bins (each bin represents a vowel and all items in a specific bin are of that vowel's type, and bins are ordered lexicographically). The condition that a string must be lexicographically sorted translates into a combinatorial problem of finding combinations with repetition.

• The problem can be approached combinatorially using:

  • The formula for combinations with repetition which is given by "n + k - 1 choose k" where n is the number of items to place, and k is the number of bins. In this context, k equals 5 (since we have five vowels), and n is the length of the string. Therefore, it becomes "n + 5 - 1 choose 5".

• This formula helps determine the number of ways to distribute n indistinguishable objects (vowels of the strings) into 5 distinguishable boxes (the vowel categories a, e, i, o, u) where the order in each box is not important, but the sequence of boxes themselves follows the lexicographical order. Thus, for any given n, this combinatory approach gives the count of valid lexicographically sorted strings.

• Example analysis:

  • For n = 2, the combinatorial logic would result in distributing two items into five bins leading to (2 + 5 - 1) choose (5) which equals 15, matching the output as explained. For n = 3, this scales accordingly, illustrating how with increasing n, the count of valid strings escalates due to the combinatorial possibilities.

Solutions

class Solution {
public:
    int calculateVowelPermutations(int k) {
        return (k + 4) * (k + 3) * (k + 2) * (k + 1) / 24;
    }
};