<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 {
    int seg[400001][3]; 
	//seg[i][0] -max length of segment in this node range, 1=lpre,
	//seg[i][1] -maximum prefix length of this segment with same character,
	//seg[i][2] -maximum suffix length of this segment with same character,
	
    string s;  // take input string as global
	
	//normal build
    void build(int i, int l, int r){
        if(l>r)return;
        if(l==r){
            seg[i][0]=1;seg[i][1]=1,seg[i][2]=1; return;
        }
        int m=(l+r)/2;
        build(2*i+1,l,m); build(2*i+2,m+1,r);
        merge(i,l,r); // merge both built child's results to this i'th node of segment tree
    }
    
    void merge(int i, int l, int r){
        int cl=2*i+1,cr=2*i+2; // left and right child segment node index
        int m=(l+r)/2;
        int cur=max(seg[cl][0],seg[cr][0]); 
        
		//initialise prefix and suffix of current node with same as prefix of left child and suffix of right child
        seg[i][1]=seg[cl][1];
        seg[i][2]=seg[cr][2];
        // if  rightmost char of left child segment ==leftmost char of right child segment
        if(s[m]==s[m+1]){
            cur=max(cur, seg[cl][2]+seg[cr][1]);
            //left= ( l to m)   ,  right= (m+1 to r)
            if(seg[cl][1]==(m-l+1)){
              seg[i][1]+=seg[cr][1];
            }
            if(seg[cr][2]==(r-(m+1)+1)){
              seg[i][2]+=seg[cl][2];
            }
        }
        seg[i][0]=cur; // store maximum of all repeating lengths
        
    }
    
    int update(int i, int l, int r, int ind,char c){
            if(ind>r or ind<l or l>r)return 0;
            if(l==r and ind==l){
                s[ind]=c;
                seg[i][0]=1;seg[i][1]=1,seg[i][2]=1; return seg[i][0];
            }
			
            int m=(l+r)/2;
            update(2*i+1,l,m,ind,c); update(2*i+2,m+1,r,ind,c);
        
            merge(i,l,r);
            return seg[i][0];
    }
    
public:
    vector<int> longestRepeating(string str, string q, vector<int>& qi) {
     
        s=str; int n=s.size();
        vector<int>ans(q.size(),1);
        build(0,0,n-1);
        for(int i=0;i<q.size();i++){
            int j=qi[i];
            char c=q[i];
            ans[i]=update(0,0,n-1,j,c);
        }
        return ans;
    }
};
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