<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 remove_seg(self, l, r):
        self.segs.remove((l, r))
        self.max_segs.remove((r-l, l, r))
        
    def add_seg(self, l, r):
        self.segs.add((l, r))
        self.max_segs.add((r-l, l, r))
        
    def longestRepeating(self, s: str, queryCharacters: str, queryIndices: List[int]) -> List[int]:
        s = list(s)
        self.segs = SortedList()
        self.max_segs = SortedList()
        l = 0
        for i in range(len(s)):
            if i and s[i] != s[i-1]:
                self.add_seg(l, i)
                l = i
        if l < len(s):
            self.add_seg(l, len(s))
        results = []
        for c, idx in zip(queryCharacters, queryIndices):
            if s[idx] == c:
                results.append(results[-1] if results else self.max_segs[-1][0])
                continue
            current = self.segs.bisect_left((idx, -inf))
            if current == len(self.segs) or self.segs[current][0] > idx:
                current -= 1
            l, r = self.segs[current]
            self.remove_seg(l, r)
            if r - (idx + 1) > 0:
                self.add_seg(idx+1, r)
            if idx - l > 0:
                self.add_seg(l, idx)
                current += 1            
            if 0 < idx < len(s) - 1 and s[idx-1] == c == s[idx+1]:
                prev = self.segs[current - 1]
                succ = self.segs[current]
                self.remove_seg(prev[0], prev[1])
                self.remove_seg(succ[0], succ[1])
                self.add_seg(prev[0], succ[1])
            elif 0 < idx and s[idx-1] == c:
                prev = self.segs[current - 1]
                self.remove_seg(prev[0], prev[1])
                self.add_seg(prev[0], idx+1)
            elif idx < len(s) - 1 and s[idx+1] == c:
                succ = self.segs[current]
                self.remove_seg(succ[0], succ[1])
                self.add_seg(idx, succ[1])
            else:
                self.add_seg(idx, idx+1)
            s[idx] = c
            results.append(self.max_segs[-1][0])
        return results
3103