<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 SegmentTree:
    def __init__(self, n, F):
        self.N = n
        self.F = F
        self.sums = [(0,0,0,0,0,0) for i in range(2*n)]
        self.dp = list(range(2*n))
        for i in range(n-1, -1, -1): self.dp[i] = self.dp[2*i]
            
        L, R = self.N, 2*self.N - 1
        parts = []
        while L <= R:
            if L & 1:
                parts += [(self.dp[L], L)]
                L += 1
            if R & 1 == 0:
                parts += [(self.dp[R], R)]
                R -= 1
            L//=2; R//=2
                
        parts.sort()
        self.d = [j for i, j in parts]
        
    def update(self, index, val):
        idx = index + self.N
        self.sums[idx] = (1, 1, 1, 1, val, val)
        while idx > 1:
            idx //= 2
            self.sums[idx] = self.F(self.sums[2*idx], self.sums[2*idx+1])
            
    def query(self):
        ans = (0, 0, 0, 0, 0, 0)
        for x in self.d:
            ans = self.F(ans, self.sums[x])
        return ans

class Solution:
    def longestRepeating(self, s, Q1, Q2):
        def FN(a, b):
            if a == (0, 0, 0, 0, 0, 0): return b
            if b == (0, 0, 0, 0, 0, 0): return a
            f1, s1, e1, L1, r1, l1 = a
            f2, s2, e2, L2, r2, l2 = b
            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(Q1)
        ans = []
        STree = SegmentTree(n, FN)
        for i in range(n):
            STree.update(i, s[i])
        for i in range(m):
            STree.update(Q2[i], Q1[i])
            ans += [STree.query()[0]]
        return ans
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