<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 typing import List, Tuple

class SegmentTree:
    def __init__(self, n: int, merge_func):
        self.n = n
        self.merge_func = merge_func
        # Initialize segment tree arrays
        self.sums = [(0, 0, 0, 0, 0, 0) for _ in range(2 * n)]
        self.dp = list(range(2 * n))
        
        # Build the segment tree
        for i in range(n - 1, -1, -1):
            self.dp[i] = self.dp[2 * i]
        
        # Collect all segment tree nodes in a specific order
        L, R = self.n, 2 * self.n - 1
        parts = []
        while L <= R:
            if L & 1:
                parts.append((self.dp[L], L))
                L += 1
            if R & 1 == 0:
                parts.append((self.dp[R], R))
                R -= 1
            L //= 2
            R //= 2
        
        parts.sort()
        self.node_order = [j for _, j in parts]
    
    def update(self, index: int, value: str):
        idx = index + self.n
        # Update leaf node
        self.sums[idx] = (1, 1, 1, 1, value, value)
        # Update parent nodes
        while idx > 1:
            idx //= 2
            self.sums[idx] = self.merge_func(self.sums[2 * idx], self.sums[2 * idx + 1])
    
    def query(self) -> Tuple[int, int, int, int, str, str]:
        result = (0, 0, 0, 0, '', '')
        for node in self.node_order:
            result = self.merge_func(result, self.sums[node])
        return result

class Solution:
    def longestRepeating(self, s: str, queryChars: str, queryIndices: List[int]) -> List[int]:
        def merge_nodes(node1, node2):
            if node1 == (0, 0, 0, 0, '', ''): return node2
            if node2 == (0, 0, 0, 0, '', ''): return node1
            
            f1, s1, e1, L1, r1, l1 = node1
            f2, s2, e2, L2, r2, l2 = node2
            
            if r1 == l2:
                return max(f1, f2, e1 + s2), s1 + s2 * (e1 == L1), e2 + e1 * (e2 == L2), L1 + L2, r2, l1
            else:
                return max(f1, f2), s1, e2, L1 + L2, r2, l1
        
        n = len(s)
        m = len(queryChars)
        result = []
        seg_tree = SegmentTree(n, merge_nodes)
        
        
        for i in range(n):
            seg_tree.update(i, s[i])
        
    
        for i in range(m):
            seg_tree.update(queryIndices[i], queryChars[i])
            result.append(seg_tree.query()[0])
        
        return result

7807