<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.l = 0
        self.r = 0
        self.lmx = 0
        self.rmx = 0
        self.mx = 0
        self.size = 0
        self.lc = None
        self.rc = None


N = 100010
tr = [Node() for _ in range(N << 2)]


class SegmentTree:
    def __init__(self, s):
        n = len(s)
        self.s = s
        self.build(1, 1, n)

    def build(self, u, l, r):
        tr[u].l = l
        tr[u].r = r
        if l == r:
            tr[u].lmx = tr[u].rmx = tr[u].mx = tr[u].size = 1
            tr[u].lc = tr[u].rc = self.s[l - 1]
            return
        mid = (l + r) >> 1
        self.build(u << 1, l, mid)
        self.build(u << 1 | 1, mid + 1, r)
        self.pushup(u)

    def modify(self, u, x, v):
        if tr[u].l == x and tr[u].r == x:
            tr[u].lc = tr[u].rc = v
            return
        mid = (tr[u].l + tr[u].r) >> 1
        if x <= mid:
            self.modify(u << 1, x, v)
        else:
            self.modify(u << 1 | 1, x, v)
        self.pushup(u)

    def query(self, u, l, r):
        if tr[u].l >= l and tr[u].r <= r:
            return tr[u]
        mid = (tr[u].l + tr[u].r) >> 1
        if r <= mid:
            return self.query(u << 1, l, r)
        if l > mid:
            return self.query(u << 1 | 1, l, r)
        left, right = self.query(u << 1, l, r), self.query(u << 1 | 1, l, r)
        res = Node()
        self._pushup(res, left, right)
        return res

    def _pushup(self, root, left, right):
        root.lc, root.rc = left.lc, right.rc
        root.size = left.size + right.size

        root.mx = max(left.mx, right.mx)
        root.lmx, root.rmx = left.lmx, right.rmx

        if left.rc == right.lc:
            if left.lmx == left.size:
                root.lmx += right.lmx
            if right.rmx == right.size:
                root.rmx += left.rmx
            root.mx = max(root.mx, left.rmx + right.lmx)

    def pushup(self, u):
        self._pushup(tr[u], tr[u << 1], tr[u << 1 | 1])


class Solution:
    def longestRepeating(
        self, s: str, queryCharacters: str, queryIndices: List[int]
    ) -> List[int]:
        tree = SegmentTree(s)
        k = len(queryIndices)
        res = []
        for i, c in enumerate(queryCharacters):
            x = queryIndices[i] + 1
            tree.modify(1, x, c)
            res.append(tree.query(1, 1, len(s)).mx)
        return res

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