# 2062. Count Vowel Substrings of a String

## Description

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') and has all five vowels present in it.

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

Example 1:

Input: word = "aeiouu"
Output: 2
Explanation: The vowel substrings of word are as follows (underlined):
- "aeiouu"
- "aeiouu"


Example 2:

Input: word = "unicornarihan"
Output: 0
Explanation: Not all 5 vowels are present, so there are no vowel substrings.


Example 3:

Input: word = "cuaieuouac"
Output: 7
Explanation: The vowel substrings of word are as follows (underlined):
- "cuaieuouac"
- "cuaieuouac"
- "cuaieuouac"
- "cuaieuouac"
- "cuaieuouac"
- "cuaieuouac"
- "cuaieuouac"


Constraints:

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

## Solutions

• class Solution {
public int countVowelSubstrings(String word) {
int n = word.length();
int ans = 0;
for (int i = 0; i < n; ++i) {
Set<Character> t = new HashSet<>();
for (int j = i; j < n; ++j) {
char c = word.charAt(j);
if (!isVowel(c)) {
break;
}
if (t.size() == 5) {
++ans;
}
}
}
return ans;
}

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

• class Solution {
public:
int countVowelSubstrings(string word) {
int ans = 0;
int n = word.size();
for (int i = 0; i < n; ++i) {
unordered_set<char> t;
for (int j = i; j < n; ++j) {
char c = word[j];
if (!isVowel(c)) break;
t.insert(c);
ans += t.size() == 5;
}
}
return ans;
}

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

• class Solution:
def countVowelSubstrings(self, word: str) -> int:
n = len(word)
s = set('aeiou')
return sum(set(word[i:j]) == s for i in range(n) for j in range(i + 1, n + 1))


• func countVowelSubstrings(word string) int {
ans, n := 0, len(word)
for i := range word {
t := map[byte]bool{}
for j := i; j < n; j++ {
c := word[j]
if !(c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u') {
break
}
t[c] = true
if len(t) == 5 {
ans++
}
}
}
return ans
}

• function countVowelSubstrings(word: string): number {
let ans = 0;
const n = word.length;
for (let i = 0; i < n; ++i) {
const t = new Set<string>();
for (let j = i; j < n; ++j) {
const c = word[j];
if (!(c === 'a' || c === 'e' || c === 'i' || c === 'o' || c === 'u')) {
break;
}