Class Solution
- java.lang.Object
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- g2201_2300.s2213_longest_substring_of_one_repeating_character.Solution
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public class Solution extends Object
2213 - Longest Substring of One Repeating Character.Hard
You are given a 0-indexed string
s
. You are also given a 0-indexed stringqueryCharacters
of lengthk
and a 0-indexed array of integer indicesqueryIndices
of lengthk
, both of which are used to describek
queries.The
ith
query updates the character ins
at indexqueryIndices[i]
to the characterqueryCharacters[i]
.Return an array
lengths
of lengthk
wherelengths[i]
is the length of the longest substring ofs
consisting of only one repeating character after theith
query is performed.Example 1:
Input: s = “babacc”, queryCharacters = “bcb”, queryIndices = [1,3,3]
Output: [3,3,4]
Explanation: - 1st query updates s = “bbbacc”. The longest substring consisting of one repeating character is “bbb” with length 3.
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2nd query updates s = “bbbccc”. The longest substring consisting of one repeating character can be “bbb” or “ccc” with length 3.
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3rd query updates s = “bbbbcc”. The longest substring consisting of one repeating character is “bbbb” with length 4.
Thus, we return [3,3,4].
Example 2:
Input: s = “abyzz”, queryCharacters = “aa”, queryIndices = [2,1]
Output: [2,3]
Explanation: - 1st query updates s = “abazz”. The longest substring consisting of one repeating character is “zz” with length 2.
- 2nd query updates s = “aaazz”. The longest substring consisting of one repeating character is “aaa” with length 3.
Thus, we return [2,3].
Constraints:
1 <= s.length <= 105
s
consists of lowercase English letters.k == queryCharacters.length == queryIndices.length
1 <= k <= 105
queryCharacters
consists of lowercase English letters.0 <= queryIndices[i] < s.length
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Constructor Summary
Constructors Constructor Description Solution()
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Method Summary
All Methods Instance Methods Concrete Methods Modifier and Type Method Description int[]
longestRepeating(String s, String queryCharacters, int[] queryIndices)
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