<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 Solution:
    def longestRepeating(self, s: str, queryCharacters: str, queryIndices: List[int]) -> List[int]:
        s = list(s)
        n = len(s)
        pre = [0] * (4 * n)
        suf = [0] * (4 * n)
        mx = [0] * (4 * n)

        def maintain(o: int, l: int, r: int) -> None:
            pre[o] = pre[o * 2]
            suf[o] = suf[o * 2 + 1]
            mx[o] = max(mx[o * 2], mx[o * 2 + 1])
            m = (l + r) // 2
            if s[m - 1] == s[m]:  # 中间字符相同，可以合并
                if suf[o * 2] == m - l + 1:
                    pre[o] += pre[o * 2 + 1]
                if pre[o * 2 + 1] == r - m:
                    suf[o] += suf[o * 2]
                mx[o] = max(mx[o], suf[o * 2] + pre[o * 2 + 1])

        def build(o: int, l: int, r: int) -> None:
            if l == r:
                pre[o] = suf[o] = mx[o] = 1
                return
            m = (l + r) // 2
            build(o * 2, l, m)
            build(o * 2 + 1, m + 1, r)
            maintain(o, l, r)

        def update(o: int, l: int, r: int, i: int) -> None:
            if l == r: return
            m = (l + r) // 2
            if i <= m:
                update(o * 2, l, m, i)
            else:
                update(o * 2 + 1, m + 1, r, i)
            maintain(o, l, r)

        build(1, 1, n)
        ans = []
        for c, i in zip(queryCharacters, queryIndices):
            s[i] = c
            update(1, 1, n, i + 1)
            ans.append(mx[1])
        return ans
4260