<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):
        self.parent = None
        self.leftChild = None
        self.rightChild = None
        self.leftIdx = None
        self.rightIdx = None
        self.leftLS = None
        self.RightLS = None
        self.LS = None

idx2leaf = {}

def build_tree(cur, leftIdx, rightIdx, s):
    if rightIdx == leftIdx:
        cur.leftIdx = cur.rightIdx = leftIdx
        cur.leftLS = 1
        cur.rightLS = 1
        cur.LS = 1
        global idx2leaf
        idx2leaf[cur.leftIdx] = cur
        return
    cur.leftIdx = leftIdx
    cur.rightIdx = rightIdx
    cur.leftChild = Node()
    cur.leftChild.parent = cur
    cur.rightChild = Node()
    cur.rightChild.parent = cur
    build_tree(cur.leftChild, leftIdx, (leftIdx+rightIdx)//2, s)
    build_tree(cur.rightChild, (leftIdx+rightIdx)//2+1, rightIdx, s)
    update_tree(cur,s)

def update_tree(cur,s):
    if s[cur.leftChild.rightIdx] != s[cur.rightChild.leftIdx]:
        cur.leftLS = cur.leftChild.leftLS
        cur.rightLS = cur.rightChild.rightLS
        cur.LS = max(cur.leftChild.LS, cur.rightChild.LS)
    else:
        if cur.leftChild.leftLS == (cur.leftChild.rightIdx - cur.leftChild.leftIdx+1):
            cur.leftLS = cur.leftChild.leftLS + cur.rightChild.leftLS
        else:
            cur.leftLS = cur.leftChild.leftLS
        if cur.rightChild.rightLS == (cur.rightChild.rightIdx - cur.rightChild.leftIdx+1):
            cur.rightLS = cur.rightChild.rightLS + cur.leftChild.rightLS
        else:
            cur.rightLS = cur.rightChild.rightLS
        cur.LS = max(cur.leftChild.rightLS+cur.rightChild.leftLS, cur.leftChild.LS, cur.rightChild.LS, cur.leftLS, cur.rightLS)

def print_tree(node,s):
    print(s[node.leftIdx:node.rightIdx+1])
    if node.leftChild:
        print_tree(node.leftChild,s)
    if node.rightChild:
        print_tree(node.rightChild,s)


class Solution:
    def longestRepeating(self, s: str, queryCharacters: str, queryIndices: List[int]) -> List[int]:
        if len(s) == 1:
            return [1]*len(queryIndices)
        s = list(s)
        root = Node()
        build_tree(root, 0, len(s)-1, s)
        global idx2leaf
        res = []
        for queryIdx, queryChar in zip(queryIndices, queryCharacters):
            # update tree
            queryNode = idx2leaf[queryIdx]
            s[queryIdx] = queryChar
            tmp = queryNode
            while tmp.parent is not None:
                update_tree(tmp.parent,s)
                tmp = tmp.parent
            res.append(root.LS)


        return res
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