LeetCode #1641 — MEDIUM

Count Sorted Vowel Strings

Move from brute-force thinking to an efficient approach using math strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math · Dynamic Programming

Example 1

Example 2

Example 3

Step 02

Core Insight

What unlocks the optimal approach

For each character, its possible values will depend on the value of its previous character, because it needs to be not smaller than it.
Think backtracking. Build a recursive function count(n, last_character) that counts the number of valid strings of length n and whose first characters are not less than last_character.
In this recursive function, iterate on the possible characters for the first character, which will be all the vowels not less than last_character, and for each possible value c, increase the answer by count(n-1, c).

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1641: Count Sorted Vowel Strings
class Solution {
    private Integer[][] f;
    private int n;

    public int countVowelStrings(int n) {
        this.n = n;
        f = new Integer[n][5];
        return dfs(0, 0);
    }

    private int dfs(int i, int j) {
        if (i >= n) {
            return 1;
        }
        if (f[i][j] != null) {
            return f[i][j];
        }
        int ans = 0;
        for (int k = j; k < 5; ++k) {
            ans += dfs(i + 1, k);
        }
        return f[i][j] = ans;
    }
}

// Accepted solution for LeetCode #1641: Count Sorted Vowel Strings
func countVowelStrings(n int) int {
	f := make([][5]int, n)
	var dfs func(i, j int) int
	dfs = func(i, j int) int {
		if i >= n {
			return 1
		}
		if f[i][j] != 0 {
			return f[i][j]
		}
		ans := 0
		for k := j; k < 5; k++ {
			ans += dfs(i+1, k)
		}
		f[i][j] = ans
		return ans
	}
	return dfs(0, 0)
}

# Accepted solution for LeetCode #1641: Count Sorted Vowel Strings
class Solution:
    def countVowelStrings(self, n: int) -> int:
        @cache
        def dfs(i, j):
            return 1 if i >= n else sum(dfs(i + 1, k) for k in range(j, 5))

        return dfs(0, 0)

// Accepted solution for LeetCode #1641: Count Sorted Vowel Strings
struct Solution;

impl Solution {
    fn count_vowel_strings(n: i32) -> i32 {
        let n = n as usize;
        let mut res = 0;
        for i in 0..5 {
            res += Self::dp(i, n);
        }
        res
    }

    fn dp(i: i32, n: usize) -> i32 {
        if n == 1 {
            1
        } else {
            let mut res = 0;
            for j in 0..=i {
                res += Self::dp(j, n - 1);
            }
            res
        }
    }
}

#[test]
fn test() {
    let n = 1;
    let res = 5;
    assert_eq!(Solution::count_vowel_strings(n), res);
    let n = 2;
    let res = 15;
    assert_eq!(Solution::count_vowel_strings(n), res);
    let n = 33;
    let res = 66045;
    assert_eq!(Solution::count_vowel_strings(n), res);
}

// Accepted solution for LeetCode #1641: Count Sorted Vowel Strings
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1641: Count Sorted Vowel Strings
// class Solution {
//     private Integer[][] f;
//     private int n;
// 
//     public int countVowelStrings(int n) {
//         this.n = n;
//         f = new Integer[n][5];
//         return dfs(0, 0);
//     }
// 
//     private int dfs(int i, int j) {
//         if (i >= n) {
//             return 1;
//         }
//         if (f[i][j] != null) {
//             return f[i][j];
//         }
//         int ans = 0;
//         for (int k = j; k < 5; ++k) {
//             ans += dfs(i + 1, k);
//         }
//         return f[i][j] = ans;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.