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2063. Vowels of All Substrings

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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.

Similar Questions:

Solution 1.

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;
    }
};

Discuss

https://leetcode.com/problems/vowels-of-all-substrings/discuss/1563756/C%2B%2B-O(N)-Time