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 <= 105wordconsists of lowercase English letters.
Similar Questions:
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/
// Author: github.com/lzl124631x
// 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;
}
};https://leetcode.com/problems/vowels-of-all-substrings/discuss/1563756/C%2B%2B-O(N)-Time