<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,SortedDict
class Solution:
    def longestRepeating(self, s: str, queryCharacters: str, queryIndices: List[int]) -> List[int]:
        n = len(s)
        lens = SortedDict() # lengths sorted order
        m = SortedList() # contain repeated intervals (start,end,char)
        m.add((-1,-1,'')) # handle edge case
        m.add((n,n,''))
        # helper functions
        def add(start,end,c): # add interval 
            m.add((start,end,c))
            lens[end-start+1] =lens.get(end-start+1,0) + 1
        def rm(ival): # remove interval
            m.remove(ival)
            ln = ival[1]-ival[0] + 1
            lens[ln] -=1
            if(lens[ln]<=0): del lens[ln]
        # building the m and lens
        i = 0
        while(i<n):
            c = s[i]
            start = i
            while(i<n and s[i]==c): i += 1
            add(start,i-1,c)
        def query(index,nc): # replace character at specific index
            pos = m.bisect_right((index,index,''))
            if(m[pos][0]>index): pos -=1 
            ival = m[pos]
            if(ival[2]==nc): return lens.peekitem()[0]; # no change if same as old character
            prev,nxt = m[pos-1],m[pos+1] # get adjacent interval for possible merge
            cur = [index,index,nc]
            if(ival[0]==index and prev[2]==nc): # try to merge with prev
                cur[0] = prev[0]
                rm(prev)
            if(ival[1]==index and nxt[2]==nc): # merge with next
                cur[1] = nxt[1]
                rm(nxt)
            rm(ival)
            if(cur): add(*cur)
            if(ival[0]<cur[0]): add(ival[0],cur[0]-1,ival[2]) # split left
            if(ival[1]>cur[1]): add(cur[1]+1,ival[1],ival[2]) # split right
            return lens.peekitem()[0]
        return [query(j,queryCharacters[i]) for i,j in enumerate(queryIndices)]
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