{ "data": { "question": { "questionId": "3208", "questionFrontendId": "2949", "categoryTitle": "Algorithms", "boundTopicId": 2538345, "title": "Count Beautiful Substrings II", "titleSlug": "count-beautiful-substrings-ii", "content": "
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 * 104
1 <= k <= 1000
s
consists of only English lowercase letters.给你一个字符串 s
和一个正整数 k
。
用 vowels
和 consonants
分别表示字符串中元音字母和辅音字母的数量。
如果某个字符串满足以下条件,则称其为 美丽字符串 :
\n\nvowels == consonants
,即元音字母和辅音字母的数量相等。(vowels * consonants) % k == 0
,即元音字母和辅音字母的数量的乘积能被 k
整除。返回字符串 s
中 非空美丽子字符串 的数量。
子字符串是字符串中的一个连续字符序列。
\n\n英语中的 元音字母 为 'a'
、'e'
、'i'
、'o'
和 'u'
。
英语中的 辅音字母 为除了元音字母之外的所有字母。
\n\n\n\n
示例 1:
\n\n\n输入:s = \"baeyh\", k = 2\n输出:2\n解释:字符串 s 中有 2 个美丽子字符串。\n- 子字符串 \"baeyh\",vowels = 2([\"a\",\"e\"]),consonants = 2([\"y\",\"h\"])。\n可以看出字符串 \"aeyh\" 是美丽字符串,因为 vowels == consonants 且 vowels * consonants % k == 0 。\n- 子字符串 \"baeyh\",vowels = 2([\"a\",\"e\"]),consonants = 2([\"b\",\"y\"])。\n可以看出字符串 \"baey\" 是美丽字符串,因为 vowels == consonants 且 vowels * consonants % k == 0 。\n可以证明字符串 s 中只有 2 个美丽子字符串。\n\n\n
示例 2:
\n\n\n输入:s = \"abba\", k = 1\n输出:3\n解释:字符串 s 中有 3 个美丽子字符串。\n- 子字符串 \"abba\",vowels = 1([\"a\"]),consonants = 1([\"b\"])。\n- 子字符串 \"abba\",vowels = 1([\"a\"]),consonants = 1([\"b\"])。\n- 子字符串 \"abba\",vowels = 2([\"a\",\"a\"]),consonants = 2([\"b\",\"b\"])。\n可以证明字符串 s 中只有 3 个美丽子字符串。\n\n\n
示例 3:
\n\n\n输入:s = \"bcdf\", k = 1\n输出:0\n解释:字符串 s 中没有美丽子字符串。\n\n\n
\n\n
提示:
\n\n1 <= s.length <= 5 * 104
1 <= k <= 1000
s
仅由小写英文字母组成。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."
],
"solution": null,
"status": null,
"sampleTestCase": "\"baeyh\"\n2",
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