longest-substring-of-one-repeating-character

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 = "bbbacc". The longest substring consisting of one repeating character is "bbb" with length 3.
- 2nd query updates s = "bbbccc". 
  The longest substring consisting of one repeating character can be "bbb" or "ccc" with length 3.
- 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