2063. Vowels of All Substrings

Problem

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 <= 10^5

• word consists of lowercase English letters.

Solution (Java)

class Solution {
    public long countVowels(String word) {
        long ans = 0;
        for (int i = 0; i < word.length(); i++) {
            if (isVowel(word.charAt(i))) {
                long right = word.length() - (long) i - 1;
                ans += ((long) i + 1) * (right + 1);
            }
        }
        return ans;
    }

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

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).