<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:
    const int n = 1e5+2;
    vector<int> a;
    vector<vector<int> > tree;

    vector<int> merge(vector<int> &left, vector<int> &right, int leftL, int leftR, int rightL, int rightR)
    {
        vector<int> newNode(6,0);

        if(left[0]==left[4] && leftR-leftL+1 == left[1] && left[4]==right[0])
        {
            newNode[0]=left[0];
            newNode[1]=left[1]+right[1];
        }
        else
        {
            newNode[0]=left[0];
            newNode[1]=left[1];
        }


        if(right[0]==right[4] && rightR-rightL+1==right[5] && left[4]==right[0])
        {
            newNode[4]=right[4];
            newNode[5]=right[5]+left[5];
        }
        else
        {
            newNode[4]=right[4];
            newNode[5]=right[5];
        }

        if(left[4]==right[0] && left[5]+right[1]>left[3] && left[5]+right[1]>right[3])
        {
            newNode[2]=left[4];
            newNode[3]=left[5]+right[1];
        }
        else
        {
            if(left[3]>right[3])
            {
                newNode[2]=left[2];
                newNode[3]=left[3];
            }
            else
            {
                newNode[2]=right[2];
                newNode[3]=right[3];
            }
        }

        return newNode;
    }

    void build(int node, int st, int en)
    {
        if(st==en)
        {
            // tree[node]=a[st];
            tree[node]=vector<int> {a[st],1,a[st],1,a[st],1};
            return;
        }
        int mid = (st+en)>>1;
        build(2*node, st, mid);
        build(2*node+1, mid+1, en);

        //The following line will depend on problem, here it stores the sum

        // tree[node] = tree[2*node]+tree[2*node+1];
        tree[node] = merge(tree[2*node], tree[2*node+1], st, mid, mid+1, en);
        return;
    }

    vector<int> query(int node, int st, int en, int l, int r)
    {
        if(r<st || l>en) return vector<int> {-1,1,-1,1,-1,1};
        if(st>=l && en<=r) return tree[node];
        int mid = (st+en)>>1;
        vector<int> q1 = query(2*node,st,mid,l,r);
        vector<int> q2 = query(2*node+1,mid+1,en,l,r);
        return merge(q1,q2,st,mid,mid+1,en);
    }

    void update(int node, int st, int en, int idx, int val)
    {
        if(st==en)
        {
            a[idx]=val;
            tree[node]=vector<int> {val,1,val,1,val,1};
            return;
        }
        int mid=(st+en)>>1;
        if(idx<=mid) update(2*node, st, mid, idx, val); 
        //<= above because mid is include in left part and right half starts from mid+1;
        else update(2*node+1, mid+1, en, idx, val);

        // tree[node] = tree[2*node]+tree[2*node+1];
        tree[node] = merge(tree[2*node],tree[2*node+1],st,mid,mid+1,en);
        return; 
    }



    vector<int> longestRepeating(string s, string queryCharacters, vector<int>& queryIndices) 
    {
        tree.assign(4*n, vector<int>());
        int l = s.length();
        a.assign(n,-1);
        for(int i=0; i<l; i++) a[i]=(int)(s[i]-'a');

        build(1,0,l-1);

        vector<int> values;
        values.reserve(n);

        int q = queryCharacters.length();
        for(int i=0; i<q; i++)
        {
            update(1,0,l-1,queryIndices[i],queryCharacters[i]-'a');
            vector<int> ans = query(1,0,l-1,0,l-1);
            values.push_back(ans[3]);
        }

        return values;
    }
};
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