<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:
string v;
struct node{
    int val,start,end,start_len,end_len;
    node*left,*right;
    node(){
        left=NULL,right=NULL;
    }
};
void create(node *root,int start,int end){
    if(start==end){
        root->val=1;
        root->start=start,root->end=end;
        root->start_len=1,root->end_len=1;
        return ;
    }
    node *left=new node(),*right=new node();
    int mid=(start+end)/2;
    create(left,start,mid);
    create(right,mid+1,end);
    root->left=left,root->right=right;
    if(v[root->left->end]==v[root->right->start])
    {
        int new_len=left->end_len+right->start_len;
        root->val=max(left->val,max(right->val,new_len));
        if(left->start_len==left->end-left->start+1)
        root->start_len=new_len;
        else
        root->start_len=left->start_len;
        if(right->end_len==right->end-right->start+1)
        root->end_len=new_len;
        else
        root->end_len=right->end_len;
        root->start=left->start;
        root->end=right->end;
    }
    else
    {
        root->val=max(left->val,right->val);
        root->start=left->start,root->end=right->end;
        root->start_len=left->start_len,root->end_len=right->end_len;
    }
    return;
    
}
int ind;
void add(node *root,int start,int end){
    if(start==end)
        return;
    int mid=(start+end)/2;
    if(ind>mid)
    add(root->right,mid+1,end);
    else
    add(root->left,start,mid);
       if(v[root->left->end]==v[root->right->start])
    {
        int new_len=root->left->end_len+root->right->start_len;
        root->val=max(root->left->val,max(root->right->val,new_len));
        if(root->left->start_len==root->left->end-root->left->start+1)
        root->start_len=new_len;
        else
        root->start_len=root->left->start_len;
        if(root->right->end_len==root->right->end-root->right->start+1)
        root->end_len=new_len;
        else
        root->end_len=root->right->end_len;
        root->start=root->left->start;
        root->end=root->right->end;
    }
    else
    {
        root->val=max(root->left->val,root->right->val);
        root->start=root->left->start,root->end=root->right->end;
        root->start_len=root->left->start_len,root->end_len=root->right->end_len;
    }
    return;
    
}
    vector<int> longestRepeating(string s, string qc, vector<int>& qi) {
       node *root=new node();
       v=s;
       create(root,0,s.size()-1);
       vector<int> res;
       for(int i=0;i<qi.size();i++){
           v[qi[i]]=qc[i];
           ind=qi[i];
           add(root,0,s.size()-1);
           res.push_back(root->val);
       }
       return res;
    }
};
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