<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>

class Solution {
public:
    vector<int> longestRepeating(string s, string qc, vector<int>& qi) {
        
        using pii = pair<int, int>;

        set<pii> ind;
        set<pii> len;
        for(int i=0;i<s.size();)
        {
            int j=i;
            while(j<s.size()&&s[j]==s[i])
                j++;
            ind.insert({i, j-1});
            len.insert({j-i, i});
            i=j;
        }
        vector<int> ans;

        for(int i=0;i<qi.size();i++)
        {
            set<pii>::iterator next = ind.upper_bound({qi[i],s.size()});
            s[qi[i]]=qc[i];
            set<pii>::iterator mid, pr; 
            mid=prev(next);
            vector<pii> tr;
            vector<pii> ta;
            int f = (*mid).first;
            int l = (*mid).second;
            if(next==ind.end() && mid==ind.begin())
            {
                tr.push_back(*mid);

                ta.push_back({f, qi[i]-1});
                ta.push_back({qi[i], qi[i]});
                ta.push_back({qi[i]+1,l});
            }
            else if(next==ind.end())
            {
                //cout<<"here2"<<endl;
                pr=prev(mid);
                tr.push_back(*mid);
                tr.push_back(*pr);

                ta.push_back(*pr);
                ta.push_back({f, qi[i]-1});
                ta.push_back({qi[i], qi[i]});
                ta.push_back({qi[i]+1,l});
                //cout<<"here2 end"<<endl;
            }
            else
            {
                pr=prev(mid);
                tr.push_back(*mid);
                tr.push_back(*pr);
                tr.push_back(*next);
                ta.push_back(*pr);
                ta.push_back({f, qi[i]-1});
                ta.push_back({qi[i], qi[i]});
                ta.push_back({qi[i]+1,l});
                ta.push_back(*next);
            }
            bool change= true;

            while(change)
            {
                change=false;
                vector<pii> temp;

                for(int i=0;i<ta.size();i++)
                {
                    if(ta[i].second<ta[i].first)
                    {
                        change=true;
                        continue;
                    }
                    temp.push_back(ta[i]);
                }
                ta = temp;
                if(change)
                    continue;
                temp.clear();
                for(int i=0;i<ta.size()-1;i++)
                {
                    int i1 = ta[i].first;
                    int i2 = ta[i+1].first;

                    //cout<<i1<<" ii "<<i2<<endl;

                    if(s[i1]==s[i2])
                    {
                        for(int j=0;j<i;j++)
                        {
                            temp.push_back(ta[j]);
                        }
                        temp.push_back({ta[i].first, ta[i+1].second});
                        for(int j=i+2;j<ta.size();j++)
                        {
                            temp.push_back(ta[j]);
                        }
                        change=true;
                        break;
                    }

                }
                if(change)
                {
                    ta=temp;
                    continue;
                }
                break;
            }
            for(int i=0;i<tr.size();i++)
            {
                //cout<<tr[i].first<<" r "<<tr[i].second<<endl;
                ind.erase(tr[i]);
                len.erase({tr[i].second-tr[i].first+1,tr[i].first});
            }
            for(int i=0;i<ta.size();i++)
            {
                //cout<<ta[i].first<<" a "<<ta[i].second<<endl;
                ind.insert(ta[i]);
                len.insert({ta[i].second-ta[i].first+1,ta[i].first});
            }

            ans.push_back((*len.rbegin()).first);
        }
        return ans;



    }
};
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