<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 SortedList

class Solution:
    def longestRepeating(self, s: str, queryCharacters: str, queryIndices: List[int]) -> List[int]:
        s = list(s)
        longest = SortedList()
        intervals = {}

        for i, char in enumerate(s):
            intervals[char] = intervals.get(char, SortedList())
            if intervals[char] and i == intervals[char][-1][1] + 1:
                intervals[char][-1][1] = i
            else:
                intervals[char].add([i, i])

        for char in intervals:
            for a, b in intervals[char]:
                longest.add(b - a + 1)

        lengths = []
        for char, index in zip(queryCharacters, queryIndices):
            old_char = s[index]
            if old_char != char:
                interval_index = intervals[old_char].bisect_left([index + 1, index + 1]) - 1
                a, b = intervals[old_char][interval_index]
                longest.remove(b - a + 1)
                intervals[old_char].remove([a, b])
                if a < index < b:
                    intervals[old_char].add([a, index - 1])
                    intervals[old_char].add([index + 1, b])
                    longest.add(index - a)
                    longest.add(b - index)
                elif index == a and a < b:
                    intervals[old_char].add([a + 1, b])
                    longest.add(b - a)
                elif index == b and a < b:
                    intervals[old_char].add([a, b - 1])
                    longest.add(b - a)
                if not intervals[old_char]:
                    del intervals[old_char]
                left = right = index
                if char in intervals:
                    interval_index = intervals[char].bisect_left([index, index]) - 1
                    if interval_index >= 0:
                        a, b = intervals[char][interval_index]
                        if b == index - 1:
                            intervals[char].remove([a, b])
                            longest.remove(b - a + 1)
                            left = a
                    if index + 1 < len(s) and s[index + 1] == char:
                        interval_index = intervals[char].bisect_left([index + 2, index + 2]) - 1
                        a, b = intervals[char][interval_index]
                        intervals[char].remove([a, b])
                        longest.remove(b - a + 1)
                        right = b
                intervals[char] = intervals.get(char, SortedList())
                intervals[char].add([left, right])
                longest.add(right - left + 1)
            lengths.append(longest[-1])
            s[index] = char

        return lengths
3200