#3541

Easy

LeetCode Problem Workspace

Find Most Frequent Vowel and Consonant

Given a string, calculate the sum of the most frequent vowel and consonant frequencies.

Problem Statement

You are given a string s consisting of lowercase English letters. Your task is to find the most frequent vowel and consonant in the string.

Return the sum of the two frequencies, where a vowel is one of 'a', 'e', 'i', 'o', 'u'. Consider only the lowercase English letters in the string.

Examples

Example 1

Input: s = "successes"

Output: 6

Example 2

Input: s = "aeiaeia"

Output: 3

Constraints

1 <= s.length <= 100
s consists of lowercase English letters only.

Solution Approach

Frequency Counting with Hashmap

The first step is to create a hashmap (or dictionary) to store the frequency of each letter. Iterate over the string, updating the hashmap with the frequency of each letter.

Identify Most Frequent Vowel and Consonant

Once the frequencies are counted, iterate through the hashmap to find the most frequent vowel and consonant. Compare the frequency of vowels with consonants and track the maximum values.

Return the Sum of Frequencies

Finally, sum the frequencies of the most frequent vowel and consonant and return this sum as the result.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of this problem depends on the approach used to iterate over the string and hashmap. Counting the frequencies of letters takes O(n) time, where n is the length of the string. Identifying the most frequent vowel and consonant takes constant time as there are only a limited number of vowels and consonants. Thus, the overall time complexity is O(n). The space complexity is O(1), assuming the hashmap size is fixed based on the limited number of possible characters in the string.

What Interviewers Usually Probe

The candidate uses a hashmap to efficiently track letter frequencies.
They properly handle both vowels and consonants in the same iteration.
The solution returns the sum of the frequencies of the most frequent vowel and consonant.

Common Pitfalls or Variants

Common pitfalls

Overcomplicating the solution by not leveraging the hashmap effectively.
Not correctly distinguishing between vowels and consonants.
Misidentifying edge cases such as strings with no vowels or consonants.

Follow-up variants

Count only vowels and consonants, ignoring other characters in the string.
Consider both uppercase and lowercase letters.
Find the sum of the most frequent letters without distinguishing between vowels and consonants.

FAQ

What is the best way to approach this problem?

The most efficient approach is to use a hashmap to track letter frequencies and then iterate through it to find the most frequent vowel and consonant.

How do I handle edge cases like strings with no vowels or consonants?

In such cases, ensure the program doesn't return incorrect results. You can handle this by initializing the frequencies of vowels and consonants to zero.

Can I use a list instead of a hashmap?

While it's possible to use a list, a hashmap is more efficient in this case as it allows direct access to the frequency of each letter, making it faster to process.

How does GhostInterview help with this problem?

GhostInterview helps by breaking down the steps needed to solve this problem, focusing on efficient data structures like hashmaps for counting frequencies.

What pattern does this problem follow?

This problem follows a 'Hash Table plus String' pattern, utilizing hashmaps for counting frequencies and efficiently identifying the most frequent vowel and consonant.

terminal

Solution

Solution 1: Counting

We first use a hash table or an array of length $26$, $\textit{cnt}$, to count the frequency of each letter. Then, we iterate through this table to find the most frequent vowel and consonant, and return the sum of their frequencies.

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class Solution:
    def maxFreqSum(self, s: str) -> int:
        cnt = Counter(s)
        a = b = 0
        for c, v in cnt.items():
            if c in "aeiou":
                a = max(a, v)
            else:
                b = max(b, v)
        return a + b

Continue Practicing

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