LeetCode 2063. Vowels of All Substrings Solution in Java, C++, Python & More | Explanation + Code

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

Solution 1: Enumerate Contribution

We can enumerate each character word[i] in the string. If word[i] is a vowel, then word[i] appears in (i + 1) × (n - i) substrings. We sum up the counts of these substrings.

The time complexity is O(n), where n is the length of the string word. The space complexity is O(1).

PythonJavaC++GoTypeScriptRustJavaScript

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')(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)

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; }(code-box)

impl Solution { pub fn count_vowels(word: String) -> i64 { let n = word.len() as i64; word.chars() .enumerate() .filter(|(_, c)| "aeiou".contains(*c)) .map(|(i, _)| (i as i64 + 1) * (n - i as i64)) .sum() } }(code-box)

/** * @param {string} word * @return {number} */ var countVowels = function (word) { 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; };(code-box)