<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, qc: str, qi):
        qr = list(zip(list(qc), qi))
        srt = SortedList([])
        mx = SortedList([])
        cr = s[0]
        sz = 1
        start, end = 0, 0
        for i in range(1, len(s)):
            if s[i] == cr:
                sz += 1
                end = i
            else:
                srt.add((start, end, sz, cr))
                mx.add(sz)
                cr = s[i]
                start = i
                end = i
                sz = 1
        srt.add((start, end, sz, cr))
        mx.add(sz)
        s = list(s)
        n = len(s)
        ans = [0] * len(qr)
        for id, (ch, ind) in enumerate(qr):
            if s[ind] == ch:
                ans[id] = mx[-1]
                continue
            s[ind] = ch

            ll = srt.bisect_left((ind, n + 1, n + 1, "z")) - 1
            start, end, sz, cho = srt[ll]
            srt.pop(ll)
            # todo start==end
            if start == end:

                mx.remove(sz)
                if ind + 1 < n and s[ind + 1] == ch and ind - 1 >= 0 and s[ind - 1] == ch:
                    ll = srt.bisect_left((ind + 1, n + 1, n + 1, "z")) - 1
                    sr, endr, szr, chr = srt[ll]
                    srt.pop(ll)
                    mx.remove(szr)
                    ll = srt.bisect_left((ind - 1, n + 1, n + 1, "z")) - 1
                    sl, endl, szl, chl = srt[ll]
                    srt.pop(ll)
                    mx.remove(szl)
                    mid = (sl, endr, szr+szl+1, ch)
                    srt.add(mid)
                    mx.add(szr+szl+1)
                elif ind + 1 < n and s[ind + 1] == ch:
                    ll = srt.bisect_left((ind + 1, n + 1, n + 1, "z")) - 1
                    sr, endr, szr, chr = srt[ll]
                    srt.pop(ll)
                    right = (ind, endr, szr + 1, chr)
                    srt.add(right)
                    mx.remove(szr)
                    mx.add(szr + 1)
                elif ind - 1 >= 0 and s[ind - 1] == ch:
                    ll = srt.bisect_left((ind - 1, n + 1, n + 1, "z")) - 1
                    sl, endl, szl, chl = srt[ll]
                    srt.pop(ll)
                    left = (sl, ind, szl + 1, ch)
                    srt.add(left)
                    mx.remove(szl)
                    mx.add(szl + 1)
                else:
                    srt.add((start, end, sz, ch))
                    mx.add(sz)

            elif ind == end:
                left = (start, end - 1, sz - 1, cho)
                srt.add(left)
                mx.remove(sz)
                mx.add(sz - 1)
                if ind + 1 < n and s[ind + 1] == ch:
                    ll = srt.bisect_left((ind + 1, n + 1, n + 1, "z")) - 1
                    sr, endr, szr, chr = srt[ll]
                    srt.pop(ll)
                    right = (ind, endr, szr + 1, chr)
                    srt.add(right)
                    mx.remove(szr)
                    mx.add(szr + 1)
                else:
                    right = (ind, ind, 1, ch)
                    srt.add(right)
                    mx.add(1)
            elif ind == start:
                right = (start + 1, end, sz - 1, cho)
                srt.add(right)
                mx.remove(sz)
                mx.add(sz-1)
                if ind - 1 >= 0 and s[ind - 1] == ch:
                    ll = srt.bisect_left((ind - 1, n + 1, n + 1, "z")) - 1
                    sl, endl, szl, chl = srt[ll]
                    srt.pop(ll)
                    left = (sl, ind, szl + 1, ch)
                    mx.remove(szl)
                    srt.add(left)
                    mx.add(szl + 1)
                else:
                    left = (ind, ind, 1, ch)
                    srt.add(left)
                    mx.add(1)
            else:
                left = (start, ind - 1, ((ind - 1)-start + 1), cho)
                mid = (ind, ind, 1, ch)
                right = (ind + 1, end, (end - (ind + 1) +1), cho)
                srt.add(left)
                srt.add(right)
                srt.add(mid)
                mx.remove(sz)
                mx.add(left[2])
                mx.add(right[2])
                mx.add(1)

            ans[id] = mx[-1]
        return ans
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