<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 heapq import heappush, heappop
class Solution:
    def longestRepeating(self, A: str, C: str, I: List[int]) -> List[int]:
        ii = [[0, 0, 1, A[0]]]
        for i in range(1, len(A)):
            if A[i] == A[i - 1]:
                ii[-1][1] = i
                ii[-1][2] += 1
            else:
                ii.append([i, i, 1, A[i]])

        hh = [-(ln) for _, _, ln, _ in ii]
        cc = defaultdict(int)
        for _, _, ln, _ in ii: cc[ln] += 1
        heapq.heapify(hh)

        rr = []
        for i in range(len(C)):
            idx = bisect_left(ii, I[i], key=lambda x: x[1])
            if ii[idx][3] == C[i]: 
                rr.append(-hh[0])
                continue
            if ii[idx][0] == ii[idx][1]:
                ii[idx][3] = C[i]
            elif ii[idx][0] == I[i]:
                ii[idx][0] = I[i] + 1
                cc[ii[idx][2]] -= 1
                ii[idx][2] -= 1
                cc[ii[idx][2]] += 1
                cc[1] += 1
                heappush(hh, -ii[idx][2])
                heappush(hh, -1)
                ii[idx:idx + 1] = [[I[i], I[i], 1, C[i]], ii[idx]]
            elif ii[idx][1] == I[i]:
                cc[ii[idx][2]] -= 1
                ii[idx][2] -= 1
                cc[ii[idx][2]] += 1
                cc[1] += 1
                heappush(hh, -ii[idx][2])
                heappush(hh, -1)
                ii[idx][1] = I[i] - 1
                ii[idx:idx + 1] = [ii[idx], [I[i], I[i], 1, C[i]]]
                idx += 1
            else:
                cc[ii[idx][2]] -= 1
                ii[idx:idx + 1] = [
                    [ii[idx][0], I[i] - 1, I[i] - ii[idx][0], ii[idx][3]],
                    [I[i], I[i], 1, C[i]], 
                    [I[i] + 1, ii[idx][1], ii[idx][1] - I[i], ii[idx][3]],
                ]
                cc[ii[idx][2]] += 1
                cc[ii[idx + 2][2]] += 1
                cc[1] += 1
                heappush(hh, -ii[idx][2])
                heappush(hh, -ii[idx + 2][2])
                heappush(hh, -1)
                idx += 1
            
            if idx + 1 < len(ii) and ii[idx + 1][3] == C[i]:
                cc[ii[idx][2]] -= 1
                cc[ii[idx + 1][2]] -= 1
                ii[idx + 1][0] = ii[idx][0]
                ii[idx + 1][2] += ii[idx][2]
                ii[idx:idx + 2] = ii[idx + 1:idx + 2]
                cc[ii[idx][2]] += 1
                heappush(hh, -ii[idx][2])
            if idx > 0 and ii[idx - 1][3] == C[i]:
                cc[ii[idx][2]] -= 1
                cc[ii[idx - 1][2]] -= 1
                ii[idx - 1][1] = ii[idx][1]
                ii[idx - 1][2] += ii[idx][2]
                ii[idx - 1:idx + 1] = ii[idx - 1:idx]
                cc[ii[idx - 1][2]] += 1
                heappush(hh, -ii[idx - 1][2])
            
            while hh and not cc[-hh[0]]:
                heappop(hh)
            rr.append(-hh[0])
        return rr
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