{ "data": { "question": { "questionId": "3223", "questionFrontendId": "2953", "categoryTitle": "Algorithms", "boundTopicId": 2548987, "title": "Count Complete Substrings", "titleSlug": "count-complete-substrings", "content": "
You are given a string word
and an integer k
.
A substring s
of word
is complete if:
s
occurs exactly k
times.2
. That is, for any two adjacent characters c1
and c2
in s
, the absolute difference in their positions in the alphabet is at most 2
.Return the number of complete substrings of word
.
A substring is a non-empty contiguous sequence of characters in a string.
\n\n\n
Example 1:
\n\n\nInput: word = "igigee", k = 2\nOutput: 3\nExplanation: The complete substrings where each character appears exactly twice and the difference between adjacent characters is at most 2 are: igigee, igigee, igigee.\n\n\n
Example 2:
\n\n\nInput: word = "aaabbbccc", k = 3\nOutput: 6\nExplanation: The complete substrings where each character appears exactly three times and the difference between adjacent characters is at most 2 are: aaabbbccc, aaabbbccc, aaabbbccc, aaabbbccc, aaabbbccc, aaabbbccc.\n\n\n
\n
Constraints:
\n\n1 <= word.length <= 105
word
consists only of lowercase English letters.1 <= k <= word.length
给你一个字符串 word
和一个整数 k
。
如果 word
的一个子字符串 s
满足以下条件,我们称它是 完全字符串:
s
中每个字符 恰好 出现 k
次。2
。也就是说,s
中两个相邻字符 c1
和 c2
,它们在字母表中的位置相差 至多 为 2
。请你返回 word
中 完全 子字符串的数目。
子字符串 指的是一个字符串中一段连续 非空 的字符序列。
\n\n\n\n
示例 1:
\n\n\n输入:word = \"igigee\", k = 2\n输出:3\n解释:完全子字符串需要满足每个字符恰好出现 2 次,且相邻字符相差至多为 2 :igigee, igigee, igigee 。\n\n\n
示例 2:
\n\n\n输入:word = \"aaabbbccc\", k = 3\n输出:6\n解释:完全子字符串需要满足每个字符恰好出现 3 次,且相邻字符相差至多为 2 :aaabbbccc, aaabbbccc, aaabbbccc, aaabbbccc, aaabbbccc, aaabbbccc 。\n\n\n
\n\n
提示:
\n\n1 <= word.length <= 105
word
只包含小写英文字母。1 <= k <= word.length
k *1, k * 2, … k * 26
.****",
"For each length, we can use sliding window to count the frequency of each letter in the window.",
"We still need to check for all characters in the window that abs(word[i] - word[i - 1]) <= 2
. We do this by maintaining the values of abs(word[i] - word[i - 1])
in the sliding window dynamically in an ordered multiset or priority queue, so that we know the maximum value at each iteration."
],
"solution": null,
"status": null,
"sampleTestCase": "\"igigee\"\n2",
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