Count redundant substrings by vowel and consonant weights

Write a function that counts the redundant substrings of a word.

A string s is redundant when its length equals a * V + b * C, where V is the number of vowels in s, C is the number of consonants in s, and a and b are given integers. The vowels are a, e, i, o, u; every other letter (including y) is a consonant.

def count_redundant_substrings(word: str, a: int, b: int) -> int: ...

Return how many of the n(n+1)/2 non-empty contiguous substrings of word are redundant. Substrings at different positions count separately, even if their text is the same.

Constraints

• 0 <= len(word) <= 10^5, lowercase English letters only.
• -10^5 <= a, b <= 10^5.
• The answer can exceed 2^31.

Example 1

word = "code", a = 2, b = 0  ->  4

The rule becomes V + C = 2V, so V == C. The redundant substrings are "co", "od", "de" and "code".

Example 2

word = "queue", a = 1, b = 0  ->  10

The rule becomes V + C = V, so C == 0. Only substrings made entirely of vowels count, and "ueue" has 10 of them.

An O(n^2) solution is too slow for the largest inputs; aim for O(n).