<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

class SegmentTree:
    def __init__(self, s: str):
        self.n = len(s)
        self.tree = [None] * (4 * self.n)
        self.build(s, 0, 0, self.n - 1)
    
    def build(self, s: str, node: int, start: int, end: int):
        if start == end:
            # 單個字符的情況
            self.tree[node] = {
                'left_char': s[start],
                'right_char': s[start],
                'left_len': 1,
                'right_len': 1,
                'max_len': 1,
                'full_len': 1
            }
        else:
            mid = (start + end) // 2
            left_child = 2 * node + 1
            right_child = 2 * node + 2
            self.build(s, left_child, start, mid)
            self.build(s, right_child, mid + 1, end)
            self.tree[node] = self.merge(self.tree[left_child], self.tree[right_child])
    
    def merge(self, left_info, right_info):
        # 合併兩個區間的信息
        result = {}
        result['left_char'] = left_info['left_char']
        result['right_char'] = right_info['right_char']
        result['left_len'] = left_info['left_len'] if left_info['full_len'] == right_info['left_len'] else left_info['left_len']
        result['right_len'] = right_info['right_len'] if right_info['full_len'] == left_info['right_len'] else right_info['right_len']
        result['max_len'] = max(left_info['max_len'], right_info['max_len'])
        if left_info['right_char'] == right_info['left_char']:
            merged_len = left_info['right_len'] + right_info['left_len']
            if merged_len > result['max_len']:
                result['max_len'] = merged_len
            if left_info['full_len'] == left_info['max_len']:
                result['left_len'] += right_info['left_len']
            if right_info['full_len'] == right_info['max_len']:
                result['right_len'] += left_info['right_len']
        result['full_len'] = left_info['full_len'] + right_info['full_len']
        return result
    
    def update(self, index: int, char: str, node: int, start: int, end: int):
        if start == end:
            self.tree[node] = {
                'left_char': char,
                'right_char': char,
                'left_len': 1,
                'right_len': 1,
                'max_len': 1,
                'full_len': 1
            }
        else:
            mid = (start + end) // 2
            left_child = 2 * node + 1
            right_child = 2 * node + 2
            if index <= mid:
                self.update(index, char, left_child, start, mid)
            else:
                self.update(index, char, right_child, mid + 1, end)
            self.tree[node] = self.merge(self.tree[left_child], self.tree[right_child])
    
    def query_max_len(self):
        return self.tree[0]['max_len']

class Solution:
    def longestRepeating(self, s: str, queryCharacters: str, queryIndices: List[int]) -> List[int]:
        seg_tree = SegmentTree(s)
        result = []
        for i in range(len(queryCharacters)):
            index = queryIndices[i]
            char = queryCharacters[i]
            seg_tree.update(index, char, 0, 0, len(s) - 1)
            result.append(seg_tree.query_max_len())
        return result

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