How do you solve Vowel-Consonant Score on LeetCode?

Vowel-Consonant Score (LeetCode #3813) 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: Vowels are defined as specific characters, while consonants are all other letters.

What is the time complexity of Vowel-Consonant Score?

The optimal solution for Vowel-Consonant Score runs in O(n) time and uses O(1) space. The complexity is O(n) as we only loop through the string once.

What is the brute force solution for Vowel-Consonant Score?

The brute force approach for Vowel-Consonant Score is Brute Force, with O(n) time complexity and O(1) space complexity. Count vowels and consonants separately by iterating through the string. Calculate the score based on the counts. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Vowel-Consonant Score?

Vowel-Consonant Score uses the following concepts: String, Simulation. The recommended patterns to study are: String Manipulation, Counting Problems.

What are the interview tips for Vowel-Consonant Score?

Clarify the definition of vowels and consonants. Discuss edge cases like empty strings or strings with only digits.

What are common mistakes when solving Vowel-Consonant Score?

Not counting spaces or digits correctly. Confusing vowels with consonants.

#3813

Vowel-Consonant Score

Easy

StringSimulationString ManipulationCounting Problems

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

Count vowels and consonants in a single pass through the string for efficiency.

⚙️

Algorithm

3 steps

1Step 1: Initialize counters for vowels and consonants.
2Step 2: Iterate through the string, updating counters based on character type.
3Step 3: Return the score based on the counts.

solution.py6 lines

1def vowel_consonant_score(s):
2    v, c = 0, 0
3    for char in s:
4        if char in 'aeiou': v += 1
5        elif char.isalpha(): c += 1
6    return v // c if c > 0 else 0

ℹ

Complexity note: The complexity is O(n) as we only loop through the string once.

1Vowels are defined as specific characters, while consonants are all other letters.
2The score calculation depends on the ratio of vowels to consonants.

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