You are given a string s and a positive integer k.
Let vowels and consonants be the number of vowels and consonants in a string.
A string is beautiful if:
\n\nvowels == consonants.(vowels * consonants) % k == 0, in other terms the multiplication of vowels and consonants is divisible by k.Return the number of non-empty beautiful substrings in the given string s.
A substring is a contiguous sequence of characters in a string.
\n\nVowel letters in English are 'a', 'e', 'i', 'o', and 'u'.
Consonant letters in English are every letter except vowels.
\n\n\n
Example 1:
\n\n\nInput: s = "baeyh", k = 2\nOutput: 2\nExplanation: There are 2 beautiful substrings in the given string.\n- Substring "baeyh", vowels = 2 (["a",e"]), consonants = 2 (["y","h"]).\nYou can see that string "aeyh" is beautiful as vowels == consonants and vowels * consonants % k == 0.\n- Substring "baeyh", vowels = 2 (["a",e"]), consonants = 2 (["b","y"]).\nYou can see that string "baey" is beautiful as vowels == consonants and vowels * consonants % k == 0.\nIt can be shown that there are only 2 beautiful substrings in the given string.\n\n\n
Example 2:
\n\n\nInput: s = "abba", k = 1\nOutput: 3\nExplanation: There are 3 beautiful substrings in the given string.\n- Substring "abba", vowels = 1 (["a"]), consonants = 1 (["b"]).\n- Substring "abba", vowels = 1 (["a"]), consonants = 1 (["b"]).\n- Substring "abba", vowels = 2 (["a","a"]), consonants = 2 (["b","b"]).\nIt can be shown that there are only 3 beautiful substrings in the given string.\n\n\n
Example 3:
\n\n\nInput: s = "bcdf", k = 1\nOutput: 0\nExplanation: There are no beautiful substrings in the given string.\n\n\n
\n
Constraints:
\n\n1 <= s.length <= 5 * 1041 <= k <= 1000s consists of only English lowercase letters.k find all the x integers such that x^2 % k == 0. Notice, that there aren’t many such candidates.",
"We can iterate over all such x values and count the number of substrings such that vowels == consonants == x.",
"This can be done with prefix sums and hash map."
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