# 2063. Vowels of All Substrings

## Description

Given a string word, return the sum of the number of vowels ('a', 'e', 'i', 'o', and 'u') in every substring of word.

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

Note: Due to the large constraints, the answer may not fit in a signed 32-bit integer. Please be careful during the calculations.

Example 1:

Input: word = "aba"
Output: 6
Explanation:
All possible substrings are: "a", "ab", "aba", "b", "ba", and "a".
- "b" has 0 vowels in it
- "a", "ab", "ba", and "a" have 1 vowel each
- "aba" has 2 vowels in it
Hence, the total sum of vowels = 0 + 1 + 1 + 1 + 1 + 2 = 6.


Example 2:

Input: word = "abc"
Output: 3
Explanation:
All possible substrings are: "a", "ab", "abc", "b", "bc", and "c".
- "a", "ab", and "abc" have 1 vowel each
- "b", "bc", and "c" have 0 vowels each
Hence, the total sum of vowels = 1 + 1 + 1 + 0 + 0 + 0 = 3.


Example 3:

Input: word = "ltcd"
Output: 0
Explanation: There are no vowels in any substring of "ltcd".


Constraints:

• 1 <= word.length <= 105
• word consists of lowercase English letters.

## Solutions

• class Solution {
public long countVowels(String word) {
long ans = 0;
for (int i = 0, n = word.length(); i < n; ++i) {
char c = word.charAt(i);
if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u') {
ans += (i + 1L) * (n - i);
}
}
return ans;
}
}

• class Solution {
public:
long long countVowels(string word) {
long long ans = 0;
for (int i = 0, n = word.size(); i < n; ++i) {
char c = word[i];
if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u') {
ans += (i + 1LL) * (n - i);
}
}
return ans;
}
};

• class Solution:
def countVowels(self, word: str) -> int:
n = len(word)
return sum((i + 1) * (n - i) for i, c in enumerate(word) if c in 'aeiou')


• func countVowels(word string) (ans int64) {
for i, c := range word {
if c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' {
ans += int64((i + 1) * (len(word) - i))
}
}
return
}

• function countVowels(word: string): number {
const n = word.length;
let ans = 0;
for (let i = 0; i < n; ++i) {
if (['a', 'e', 'i', 'o', 'u'].includes(word[i])) {
ans += (i + 1) * (n - i);
}
}
return ans;
}