<p>You are given a <strong>0-indexed</strong> string <code>s</code>. You are also given a <strong>0-indexed</strong> string <code>queryCharacters</code> of length <code>k</code> and a <strong>0-indexed</strong> array of integer <strong>indices</strong> <code>queryIndices</code> of length <code>k</code>, both of which are used to describe <code>k</code> queries.</p>

<p>The <code>i<sup>th</sup></code> query updates the character in <code>s</code> at index <code>queryIndices[i]</code> to the character <code>queryCharacters[i]</code>.</p>

<p>Return <em>an array</em> <code>lengths</code> <em>of length </em><code>k</code><em> where</em> <code>lengths[i]</code> <em>is the <strong>length</strong> of the <strong>longest substring</strong> of </em><code>s</code><em> consisting of <strong>only one repeating</strong> character <strong>after</strong> the</em> <code>i<sup>th</sup></code> <em>query</em><em> is performed.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;babacc&quot;, queryCharacters = &quot;bcb&quot;, queryIndices = [1,3,3]
<strong>Output:</strong> [3,3,4]
<strong>Explanation:</strong> 
- 1<sup>st</sup> query updates s = &quot;<u>b<strong>b</strong>b</u>acc&quot;. The longest substring consisting of one repeating character is &quot;bbb&quot; with length 3.
- 2<sup>nd</sup> query updates s = &quot;bbb<u><strong>c</strong>cc</u>&quot;. 
  The longest substring consisting of one repeating character can be &quot;bbb&quot; or &quot;ccc&quot; with length 3.
- 3<sup>rd</sup> query updates s = &quot;<u>bbb<strong>b</strong></u>cc&quot;. The longest substring consisting of one repeating character is &quot;bbbb&quot; with length 4.
Thus, we return [3,3,4].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abyzz&quot;, queryCharacters = &quot;aa&quot;, queryIndices = [2,1]
<strong>Output:</strong> [2,3]
<strong>Explanation:</strong>
- 1<sup>st</sup> query updates s = &quot;ab<strong>a</strong><u>zz</u>&quot;. The longest substring consisting of one repeating character is &quot;zz&quot; with length 2.
- 2<sup>nd</sup> query updates s = &quot;<u>a<strong>a</strong>a</u>zz&quot;. The longest substring consisting of one repeating character is &quot;aaa&quot; with length 3.
Thus, we return [2,3].
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
	<li><code>k == queryCharacters.length == queryIndices.length</code></li>
	<li><code>1 &lt;= k &lt;= 10<sup>5</sup></code></li>
	<li><code>queryCharacters</code> consists of lowercase English letters.</li>
	<li><code>0 &lt;= queryIndices[i] &lt; s.length</code></li>
</ul>

from sortedcontainers import SortedList

class Solution:
    def longestRepeating(self, s: str, queryCharacters: str, queryIndices: List[int]) -> List[int]:
        sl = SortedList()
        length = SortedList()
        curr = 0
        for char, it in itertools.groupby(s):
            c = sum(1 for _ in it)
            length.add(c)
            sl.add((curr, curr + c, char))
            curr += c
        ans = []
        for char, i in zip(queryCharacters, queryIndices):
            t = (i, math.inf, 'a')
            index = sl.bisect_right(t) - 1
            to_remove = [sl[index]]
            to_add = []
            left, right, original_char = to_remove[0]
            if original_char != char:
                length.remove(right - left)
                if right - left > 1:
                    if i == left:
                        left += 1
                        to_add.append((left, right, original_char))
                        length.add(right - left)
                    elif i == right - 1:
                        right -= 1
                        to_add.append((left, right, original_char))
                        length.add(right - left)
                    else:
                        to_add.append((left, i, original_char))
                        length.add(i - left)
                        to_add.append((i + 1, right, original_char))
                        length.add(right - (i + 1))
                
                l, r = i, i + 1
                if index - 1 >= 0 and sl[index - 1][1:3] == (i, char):
                    l, old_r, _ = sl[index - 1]
                    to_remove.append(sl[index - 1])
                    length.remove(old_r - l)
                if index + 1 < len(sl) and sl[index + 1][0] == i + 1 and sl[index + 1][2] == char:
                    old_l, r, old_length = sl[index + 1]
                    to_remove.append(sl[index + 1])
                    length.remove(r - old_l)
                length.add(r - l)
                sl.add((l, r, char))
                for t in to_remove:
                    sl.remove(t)
                sl.update(to_add)
            # print(sl)
            # print(length)
            ans.append(length[-1])

        return ans
3440