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** string `queryCharacters` of length `k` and a **0-indexed** array of integer **indices** `queryIndices` of length `k`, both of which are used to describe `k` queries.
The ith
query updates the character in `s` at index `queryIndices[i]` to the character `queryCharacters[i]`.
Return _an array_ `lengths` _of length_ `k` _where_ `lengths[i]` _is the **length** of the **longest substring** of_ `s` _consisting of **only one repeating** character **after** the_ ith
_query_ _is performed._
**Example 1:**
**Input:** s = "babacc", queryCharacters = "bcb", queryIndices = [1,3,3]
**Output:** [3,3,4]
**Explanation:** - 1st query updates s = "b**b**bacc". The longest substring consisting of one repeating character is "bbb" with length 3.
- 2nd query updates s = "bbb**c**cc". The longest substring consisting of one repeating character can be "bbb" or "ccc" with length 3.
- 3rd query updates s = "bbb**b**cc". 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 = "ab**a**zz". The longest substring consisting of one repeating character is "zz" with length 2.
- 2nd query updates s = "a**a**azz". 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`