Count Vowel Substrings of a String - Practice Coding Problems

Count Vowel Substrings of a String - Problem

A substring is a contiguous (non-empty) sequence of characters within a string.

A vowel substring is a substring that:

Only consists of vowels ('a', 'e', 'i', 'o', and 'u')
Has all five vowels present in it

Given a string word, return the number of vowel substrings in word.

Input & Output

Example 1 — Basic Case

$ Input: word = "aeiouu"

› Output: 2

💡 Note: The vowel substrings with all 5 vowels are: "aeiou" (positions 0-4) and "aeiouu" (positions 0-5). Both contain all vowels a,e,i,o,u.

Example 2 — With Consonants

$ Input: word = "unicornarihan"

› Output: 0

💡 Note: No substring contains all 5 vowels. The vowel-only segments are too short: "u", "i", "o", "a", "i", "a" - none have all 5 vowels.

Example 3 — Mixed Content

$ Input: word = "cuaieuouuc"

› Output: 7

💡 Note: The vowel segment "uaieuous" (positions 1-7) contains all 5 vowels. Valid substrings are: "uaieuo", "uaieuou", "uaieuouu", "aieuou", "aieuouu", "ieuouu", "aieuouu" - total of 7.

Constraints

1 ≤ word.length ≤ 100
word consists of lowercase English letters only

Visualization

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Understanding the Visualization

Input Analysis

String "cuaieuouuc" with vowels and consonants

Find Vowel Segments

Identify continuous vowel-only sections

Count Valid Substrings

Count substrings with all 5 unique vowels

Key Takeaway

🎯 Key Insight: Valid substrings need exactly 5 unique vowels and no consonants

Asked in

G Google 12 a Amazon 8 M Microsoft 6 f Facebook 4

The key insight is that a valid vowel substring must contain all 5 vowels (a,e,i,o,u) and no consonants. The optimal approach uses sliding window to efficiently track vowel counts and avoid redundant validation. Time: O(n²), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Sliding Window with Counting	O(n²)	O(1)	Use sliding window to efficiently count vowel substrings by tracking when all 5 vowels are present
Optimized - Skip Invalid Sections	O(n³)	O(1)	Skip consonant sections and only check vowel-only segments for all 5 vowels
Brute Force - Check All Substrings	O(n³)	O(1)	Check every possible substring to see if it contains all 5 vowels and no consonants

Sliding Window with Counting — Algorithm Steps

Identify vowel-only segments
For each start position, extend window until 5 vowels found
Count all valid endings from that point
Move to next start position

Visualization

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Step-by-Step Walkthrough

Start Window

Begin at each vowel position

Extend Until Complete

Grow window until all 5 vowels found

Count Extensions

Count all valid extensions from completion point

Code -

solution.c — C

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

int getVowelIndex(char c) {
    if (c == 'a') return 0;
    if (c == 'e') return 1;
    if (c == 'i') return 2;
    if (c == 'o') return 3;
    if (c == 'u') return 4;
    return -1;
}

int solution(char* word) {
    int count = 0;
    int n = strlen(word);
    
    for (int i = 0; i < n; i++) {
        int vowelIdx = getVowelIndex(word[i]);
        if (vowelIdx == -1) {
            continue;
        }
        
        int vowelCount[5] = {0};
        int uniqueVowels = 0;
        
        for (int j = i; j < n; j++) {
            vowelIdx = getVowelIndex(word[j]);
            if (vowelIdx == -1) {
                break;
            }
            
            if (vowelCount[vowelIdx] == 0) {
                uniqueVowels++;
            }
            vowelCount[vowelIdx]++;
            
            if (uniqueVowels == 5) {
                // Count all valid endings from this point
                int k = j;
                while (k < n && getVowelIndex(word[k]) != -1) {
                    k++;
                }
                count += k - j;
                break;
            }
        }
    }
    
    return count;
}

int main() {
    char word[1000];
    fgets(word, sizeof(word), stdin);
    word[strcspn(word, "\n")] = 0;
    
    int result = solution(word);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Outer loop O(n) starting positions, inner loop O(n) to find all 5 vowels

⚠ Quadratic Growth

Space Complexity

O(1)

Only using fixed-size arrays to track vowel counts

✓ Linear Space

23.4K Views

Medium Frequency

~25 min Avg. Time

856 Likes

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Sliding Window with Counting — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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