Count Vowel Substrings of a String

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: 6

💡 Note: The vowel segment "uaieuouu" (positions 1-8) contains all 5 vowels. Valid substrings with all 5 vowels are: "uaieuo" (1-6), "uaieuou" (1-7), "uaieuouu" (1-8), "aieuo" (2-6), "aieuou" (2-7), "aieuouu" (2-8) - total of 6.

Constraints

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

Visualization

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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

✓ Brute Force - Check All Substrings

⏱️ Time: O(n³) Space: O(1)

Generate all possible substrings, then for each substring check if it contains only vowels and has all 5 vowels present. Count the valid ones.

Optimized - Skip Invalid Sections

⏱️ Time: O(n³) Space: O(1)

Instead of checking every substring, find continuous vowel-only segments first, then only check substrings within these segments. This avoids wasting time on substrings containing consonants.

Sliding Window with Counting

⏱️ Time: O(n²) Space: O(1)

For each starting position in vowel segments, extend the window until all 5 vowels are found, then count all valid endings. This avoids redundant validation of the same prefixes.

Brute Force - Check All Substrings — Algorithm Steps

Generate all possible substrings
For each substring, check if it contains only vowels
Check if all 5 vowels are present
Count valid substrings

Visualization

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Step-by-Step Walkthrough

Generate Substrings

Create all possible contiguous substrings

Validate Each

Check if contains only vowels and all 5 vowels

Count Valid

Increment counter for valid vowel substrings

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* word) {
    int count = 0;
    int n = strlen(word);
    
    for (int i = 0; i < n; i++) {
        for (int j = i; j < n; j++) {
            // Check if substring contains only vowels and all 5 vowels
            bool onlyVowels = true;
            bool hasA = false, hasE = false, hasI = false, hasO = false, hasU = false;
            
            for (int k = i; k <= j; k++) {
                if (!isVowel(word[k])) {
                    onlyVowels = false;
                    break;
                }
                if (word[k] == 'a') hasA = true;
                else if (word[k] == 'e') hasE = true;
                else if (word[k] == 'i') hasI = true;
                else if (word[k] == 'o') hasO = true;
                else if (word[k] == 'u') hasU = true;
            }
            
            if (onlyVowels && hasA && hasE && hasI && hasO && hasU) {
                count++;
            }
        }
    }
    
    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(n³)

O(n²) substrings, each taking O(n) time to validate

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables to track vowel counts

✓ Linear Space

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Medium Frequency

~25 min Avg. Time

856 Likes

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Input & Output

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Related Problems

Common Approaches

Brute Force - Check All Substrings — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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