Formatted question description: https://leetcode.ca/all/2063.html

# 2063. Vowels of All Substrings (Medium)

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".

Example 4:

Input: word = "noosabasboosa"
Output: 237
Explanation: There are a total of 237 vowels in all the substrings.


Constraints:

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

Similar Questions:

## Solution 1.

If s[i] is vowel, there are (i + 1) * (N - i) substrings that contain s[i] where i + 1 and N - i are the number of possible left and right end points of the substrings, respectively.

// OJ: https://leetcode.com/problems/vowels-of-all-substrings/
// Time: O(N)
// Space: O(1)
class Solution {
bool isVowel(char c) {
return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
};
public:
long long countVowels(string s) {
long long N = s.size(), ans = 0;
for (long long i = 0; i < N; ++i) {
if (!isVowel(s[i])) continue;
ans += (i + 1) * (N - i);
}
return ans;
}
};


## Discuss

https://leetcode.com/problems/vowels-of-all-substrings/discuss/1563756/C%2B%2B-O(N)-Time