Problem Description

The "Count Sorted Vowel Strings" problem asks you to determine how many strings of length n can be formed using only the vowels 'a', 'e', 'i', 'o', and 'u', such that the string is sorted lexicographically (each character is the same as or comes after the previous one in the alphabet).

Constraints:

• Each string must be exactly n characters long.
• Characters can be repeated.
• The characters in the string must be in non-decreasing (sorted) order.
• Input: n (an integer, 1 ≤ n ≤ 50)
• Output: The total number of valid strings.

Thought Process

At first glance, one might consider generating all possible strings of length n using the five vowels and then filtering those that are sorted. However, this brute-force approach quickly becomes infeasible as n grows, since there are 5^n possible strings.

To optimize, we recognize that the problem is about counting combinations, not generating them. Since the order of vowels in the string must be non-decreasing, this is akin to distributing n indistinguishable balls (positions in the string) into 5 distinguishable bins (vowels, in sorted order), where each bin can hold zero or more balls. This is a classic combinatorics problem known as "stars and bars."

Solution Approach

Let's break down the algorithm for counting sorted vowel strings:

1. Recognize the Combinatorial Structure:
   • We need to count the number of ways to place n vowels (with repetition) into 5 ordered slots (the vowels 'a', 'e', 'i', 'o', 'u').
   • This is equivalent to the problem of counting the number of multisets of size n from a set of 5 items.
2. Apply the "Stars and Bars" Theorem:
   • The number of multisets of size n from 5 types is C(n + 5 - 1, 5 - 1), or C(n+4, 4), where C denotes the binomial coefficient ("n choose k").
3. Efficient Calculation:
   • Calculate the binomial coefficient C(n+4, 4) efficiently using either a loop, a helper function, or built-in math libraries.
4. Return the Result:
   • Return the computed value as the answer.

This approach is highly efficient because it avoids generating any strings and relies purely on mathematical calculation.

Example Walkthrough

Let's walk through an example with n = 2:

• We want to count all sorted strings of length 2 made from 'a', 'e', 'i', 'o', 'u'.
• By hand, these are:
  • aa, ae, ai, ao, au
  • ee, ei, eo, eu
  • ii, io, iu
  • oo, ou
  • uu
  That's 15 strings.
• Using the formula: C(n+4, 4) = C(6, 4) = 15.
• The formula matches the manual count, confirming correctness.

Time and Space Complexity

• Brute-force approach:
  • Time Complexity: O(5^n) (generate all strings, check if sorted)
  • Space Complexity: Potentially O(5^n) if storing all strings
• Optimized combinatorial approach:
  • Time Complexity: O(1) (just compute a binomial coefficient)
  • Space Complexity: O(1)
• The optimized approach is vastly superior, especially for large n.

Summary

The "Count Sorted Vowel Strings" problem is a classic example of reducing a seemingly complex enumeration task to a simple combinatorial formula. By recognizing the connection to the "stars and bars" theorem, we can compute the result in constant time, regardless of the input size. This shift from brute-force to mathematical reasoning is a key insight in algorithm design, making the solution both elegant and efficient.

Code Implementation

import math

class Solution:
    def countVowelStrings(self, n: int) -> int:
        # Using stars and bars: C(n+4, 4)
        return math.comb(n + 4, 4)
      

class Solution {
public:
    int countVowelStrings(int n) {
        // C(n+4, 4) = (n+4)*(n+3)*(n+2)*(n+1)/24
        return (n+4)*(n+3)*(n+2)*(n+1)/24;
    }
};
      

class Solution {
    public int countVowelStrings(int n) {
        // C(n+4, 4) = (n+4)*(n+3)*(n+2)*(n+1)/24
        return (n+4)*(n+3)*(n+2)*(n+1)/24;
    }
}
      

var countVowelStrings = function(n) {
    // C(n+4, 4) = (n+4)*(n+3)*(n+2)*(n+1)/24
    return ((n+4)*(n+3)*(n+2)*(n+1))/24;
};