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Java-based LeetCode algorithm problem solutions, regularly updated

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package g2001_2100.s2063_vowels_of_all_substrings;

// #Medium #String #Dynamic_Programming #Math #Combinatorics
// #2022_05_29_Time_21_ms_(34.49%)_Space_48.9_MB_(68.06%)

/**
 * 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".
 *
 * **Constraints:**
 *
 * *   1 <= word.length <= 10⁵
 * *   `word` consists of lowercase English letters.
**/
public 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';
    }
}