Find Most Frequent Vowel and Consonant - Practice Coding Problems

Find Most Frequent Vowel and Consonant - Problem

You are working as a linguistic analyst for a language learning app. Given a string s consisting of lowercase English letters, your task is to identify the most popular vowel and consonant characters.

Your mission:

Find the vowel (one of 'a', 'e', 'i', 'o', or 'u') with the maximum frequency
Find the consonant (all other letters excluding vowels) with the maximum frequency
Return the sum of these two maximum frequencies

Special cases:

If multiple vowels or consonants tie for the highest frequency, you can choose any one of them
If there are no vowels or no consonants in the string, consider their frequency as 0

Example: In the string "programming", the vowel 'o' appears 1 time and consonant 'r' appears 2 times, so we return 1 + 2 = 3.

Input & Output

example_1.py — Basic case

$ Input: s = "programming"

› Output: 3

💡 Note: Vowels: 'o' appears 1 time, 'a' and 'i' appear 1 time each. Max vowel frequency = 1. Consonants: 'r' appears 2 times, 'm' appears 2 times, others appear 1 time each. Max consonant frequency = 2. Result: 1 + 2 = 3.

example_2.py — Equal frequencies

$ Input: s = "aabbcc"

› Output: 4

💡 Note: Vowels: 'a' appears 2 times. Max vowel frequency = 2. Consonants: 'b' appears 2 times, 'c' appears 2 times. Max consonant frequency = 2. Result: 2 + 2 = 4.

example_3.py — Only vowels

$ Input: s = "aeiou"

› Output: 1

💡 Note: Vowels: Each vowel 'a', 'e', 'i', 'o', 'u' appears 1 time. Max vowel frequency = 1. Consonants: None present. Max consonant frequency = 0. Result: 1 + 0 = 1.

Constraints

1 ≤ s.length ≤ 1000
s consists of lowercase English letters only
Vowels are defined as 'a', 'e', 'i', 'o', 'u'
All other lowercase letters are consonants

Visualization

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Understanding the Visualization

Initialize Trackers

Set up frequency counters and maximum trackers for both vowels and consonants

Process Each Character

For each character, increment its frequency and update the appropriate maximum

Categorize Efficiently

Use a vowel set for O(1) lookup to determine if character is vowel or consonant

Return Sum

Add the maximum vowel frequency and maximum consonant frequency

Key Takeaway

🎯 Key Insight: By tracking maximums during frequency counting, we eliminate the need for a separate search phase, achieving optimal O(n) performance.

Asked in

G Google 15 a Amazon 23 f Meta 8 ⊞ Microsoft 12

The optimal solution uses a one-pass hash table approach with O(n) time complexity. As we iterate through the string, we simultaneously count character frequencies and track maximum frequencies for vowels and consonants. This eliminates the need for multiple passes or expensive search operations, making it both time and space efficient.

Common Approaches

Approach	Time	Space	Notes
✓ One-Pass Hash Table (Optimal)	O(n)	O(1)	Count all character frequencies in one pass while tracking maximums
Brute Force (Nested Character Counting)	O(n × 26)	O(1)	Count frequency of each possible character by scanning the entire string

One-Pass Hash Table (Optimal) — Algorithm Steps

Initialize counters for max vowel and consonant frequencies
Create a set of vowels for quick lookup
Iterate through each character in the string once
For each character, increment its frequency count
Simultaneously update the appropriate maximum (vowel or consonant)
Return the sum of both maximums

Visualization

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Step-by-Step Walkthrough

Process 'p'

Consonant 'p': freq=1, max_consonant=1

Process 'r'

Consonant 'r': freq=1, max_consonant=1

Process 'o'

Vowel 'o': freq=1, max_vowel=1

Continue efficiently

Single pass through all characters

Code -

solution.c — C

#include <stdio.h>
#include <string.h>
#include <stdbool.h>

bool isVowel(char c) {
    return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
}

int solution(char* s) {
    int charFreq[26] = {0}; // Frequency array for a-z
    int maxVowelFreq = 0;
    int maxConsonantFreq = 0;
    int len = strlen(s);
    
    // Single pass through the string
    for (int i = 0; i < len; i++) {
        char ch = s[i];
        int index = ch - 'a';
        
        // Update frequency
        charFreq[index]++;
        int currentFreq = charFreq[index];
        
        // Update appropriate maximum
        if (isVowel(ch)) {
            if (currentFreq > maxVowelFreq) {
                maxVowelFreq = currentFreq;
            }
        } else {
            if (currentFreq > maxConsonantFreq) {
                maxConsonantFreq = currentFreq;
            }
        }
    }
    
    return maxVowelFreq + maxConsonantFreq;
}

int main() {
    char s[1001];
    scanf("%s", s);
    // Remove quotes if present
    if (s[0] == '"') {
        int len = strlen(s);
        for (int i = 1; i < len - 1; i++) {
            s[i-1] = s[i];
        }
        s[len-2] = '\0';
    }
    printf("%d\n", solution(s));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

We scan through the string exactly once, where n is the length of the string

✓ Linear Growth

Space Complexity

O(1)

We use at most 26 entries in our frequency map (constant space) plus a few variables

✓ Linear Space

18.2K Views

Medium Frequency

~12 min Avg. Time

892 Likes

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

One-Pass Hash Table (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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