<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 sortedcontainers import SortedDict

class Solution:
    def longestRepeating(self, s: str, queryCharacters: str, queryIndices: List[int]) -> List[int]:
        n = len(s)
        sd = SortedDict()
        i = 0
        lengths = []
        s = list(s)
        while i < n:
            old_i = i
            i += 1
            while i < n and s[i] == s[old_i]:
                i += 1
            l = i-old_i
            sd[old_i] = l
            lengths.append((-l, old_i))
        heapify(lengths)

        def update_heap():
            while lengths and (lengths[0][1] not in sd or sd[lengths[0][1]] != -lengths[0][0]):
                heappop(lengths)

        def split_at(i):
            # print("  xx split at", i, sd)
            start, width = sd.items()[sd.bisect_right(i-1)-1]
            sd[start] = i - start
            sd[i] = start+width - i
            heappush(lengths, (-sd[start], start))
            heappush(lengths, (-sd[i], i))
        
        def merge_at(i):
            # print("  xx merge at", i, sd)
            width = sd[i]
            # print("bl", sd.bisect_right(i-1)-1)
            start = sd.keys()[sd.bisect_right(i-1)-1]
            # print("start", start)
            del sd[i]
            sd[start] += width
            heappush(lengths, (-sd[start], start))
        
        res = []
        for qi, qc in zip(queryIndices, queryCharacters):
            # print("qi/qc", qi, qc)
            if s[qi] != qc:
                if qi+1 < n and s[qi+1] == qc:
                    merge_at(qi+1)
                elif qi+1 < n and s[qi+1] == s[qi]:
                    split_at(qi+1)
                
                if qi > 0 and s[qi-1] == qc:
                    merge_at(qi)
                elif qi > 0 and s[qi-1] == s[qi]:
                    split_at(qi)
                
                update_heap()
                s[qi] = qc
                # print("-->", sd)

            res.append(-lengths[0][0])
        return res
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