<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>

class Node:
    def __init__(self, start, end):
        self.start = start
        self.end = end
        self.left = self.right = None
        self.max_len = self.prefix = self.suffix = 0
        self.left_char = self.right_char = None

class SegmentTree:
    def __init__(self, s):
        self.root = self._build(s, 0, len(s) - 1)

    def _build(self, s, start, end):
        if start > end:
            return None
        
        node = Node(start, end)
        
        if start == end:
            node.max_len = node.prefix = node.suffix = 1
            node.left_char = node.right_char = s[start]
            return node
        
        mid = (start + end) // 2
        node.left = self._build(s, start, mid)
        node.right = self._build(s, mid + 1, end)
        
        self._update_node(node)
        return node

    def _update_node(self, node):
        if not node.left and not node.right:
            return

        node.left_char = node.left.left_char
        node.right_char = node.right.right_char
        
        node.prefix = node.left.prefix
        if node.left.prefix == node.left.end - node.left.start + 1 and node.left.right_char == node.right.left_char:
            node.prefix += node.right.prefix
        
        node.suffix = node.right.suffix
        if node.right.suffix == node.right.end - node.right.start + 1 and node.left.right_char == node.right.left_char:
            node.suffix += node.left.suffix
        
        node.max_len = max(node.left.max_len, node.right.max_len)
        if node.left.right_char == node.right.left_char:
            node.max_len = max(node.max_len, node.left.suffix + node.right.prefix)

    def update(self, index, char):
        self._update(self.root, index, char)

    def _update(self, node, index, char):
        if node.start == node.end == index:
            node.left_char = node.right_char = char
            return
        
        mid = (node.start + node.end) // 2
        if index <= mid:
            self._update(node.left, index, char)
        else:
            self._update(node.right, index, char)
        
        self._update_node(node)

    def query(self):
        return self.root.max_len

class Solution:
    def longestRepeating(self, s: str, queryCharacters: str, queryIndices: List[int]) -> List[int]:
        tree = SegmentTree(s)
        result = []
        
        for char, index in zip(queryCharacters, queryIndices):
            tree.update(index, char)
            result.append(tree.query())
        
        return result
4733