How do you solve Count Sorted Vowel Strings on LeetCode?

Count Sorted Vowel Strings (LeetCode #1641) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding combinatorial counting can simplify problems significantly.

What is the brute force solution for Count Sorted Vowel Strings?

The brute force approach for Count Sorted Vowel Strings is Brute Force, with O(n²) time complexity and O(1) space complexity. We can generate all possible combinations of vowel strings of length n and then filter out the ones that are lexicographically sorted. This approach is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Sorted Vowel Strings?

Count Sorted Vowel Strings uses the following concepts: Math, Dynamic Programming, Combinatorics. The recommended patterns to study are: Combinatorics, Dynamic Programming.

What are the interview tips for Count Sorted Vowel Strings?

Always start with a brute-force solution to understand the problem better. Look for patterns or mathematical properties that can simplify your solution. Be prepared to explain your thought process and why you chose a particular approach.

What are common mistakes when solving Count Sorted Vowel Strings?

Generating all combinations without considering efficiency. Not recognizing the combinatorial nature of the problem.

#1641

Count Sorted Vowel Strings

Medium

MathDynamic ProgrammingCombinatoricsCombinatoricsDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

Instead of generating all strings, we can use combinatorial mathematics. The problem can be reduced to counting combinations of vowels with repetitions allowed, which can be calculated using the formula for combinations.

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Algorithm

3 steps

1Step 1: Recognize that the problem can be represented as choosing n items from 5 types (vowels) with repetition allowed.
2Step 2: Use the combinatorial formula C(n + k - 1, k - 1) where n is the length of the string and k is the number of vowels.
3Step 3: Calculate the result using factorials or dynamic programming.

solution.py5 lines

1# Full working Python code
2from math import comb
3
4def countVowelStrings(n):
5    return comb(n + 4, 4)

ℹ

Complexity note: The time complexity is O(n) due to the loop in the combination calculation, while space complexity is O(1) since we only use a few variables.

1Understanding combinatorial counting can simplify problems significantly.
2Recognizing patterns in the problem can lead to more efficient solutions.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.