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.
class Solution:
def countVowels(self, word: str) -> int:
ans, n = 0, len(word)
for i, c in enumerate(word):
if c in ['a', 'e', 'i', 'o', 'u']:
ans += (i + 1) * (n - i)
return ans
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 += (long) (i + 1) * (n - i);
}
}
return ans;
}
}
function countVowels(word: string): number {
const n = word.length;
let ans = 0;
for (let i = 0; i < n; i++) {
let char = word.charAt(i);
if (['a', 'e', 'i', 'o', 'u'].includes(char)) {
ans += ((i + 1) * (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 += (long long) (i + 1) * (n - i);
}
return ans;
}
};
func countVowels(word string) int64 {
var ans int64
n := len(word)
for i, c := range word {
if c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' {
ans += int64((i + 1) * (n - i))
}
}
return ans
}