<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
{
private:
    struct Node
    {
        int max, size;
        int prefix_size, suffix_size;
        char prefix_char, suffix_char;
        Node(char ch)
        {
            max = size = 1;
            prefix_size = suffix_size = 1;
            prefix_char = suffix_char = ch;
        }
    };

    string a;
    vector<Node> tree;
    
    Node merge_node(Node left, Node right)
    {
        Node ans('a');

        // merge prefix
        ans.prefix_size = left.prefix_size;
        
        if(left.prefix_size == left.size && left.prefix_char == right.prefix_char)
            ans.prefix_size = left.prefix_size + right.prefix_size;
        
        ans.prefix_char = left.prefix_char;
        
        // merge suffix
        ans.suffix_size = right.suffix_size;
        
        if(right.suffix_size == right.size && right.suffix_char == left.suffix_char)
            ans.suffix_size = right.suffix_size + left.suffix_size;
        
        ans.suffix_char = right.suffix_char;
        
        // size of this segment node
        ans.size = left.size + right.size;
        
        // max size
        int tmp = max(ans.suffix_size, ans.prefix_size);
        
        if(left.suffix_char == right.prefix_char)
            tmp = max(tmp, left.suffix_size + right.prefix_size);
        
        ans.max = max({left.max, right.max, tmp});
        
        return ans;
    }
    
    Node buildST(int i, int l, int r)
    {
        if(l == r)
        {
            Node node(a[l]);
            tree[i] = node;
            return tree[i];
        }

        int mid = (l + r) / 2;
        Node left = buildST(2*i+1, l, mid);
        Node right = buildST(2*i+2, mid+1, r);
    
        return tree[i] = merge_node(left, right);
    }
    
    void update(int i, int sl, int sr, int pos, char ch)
    {
        if(sl == sr)
        {
            //update in the leaf node of the tree
            Node node(ch);
            tree[i] = node;
            return;
        }

        int mid = (sl + sr) / 2;
        if(pos <= mid) 
            update(2*i+1, sl, mid, pos, ch);

        else 
            update(2*i+2, mid+1, sr, pos, ch);

        //update intermediate nodes after returning the value of the leaf node
        tree[i] = merge_node(tree[2*i+1], tree[2*i+2]);      
    }    
public:
    vector<int> longestRepeating(string s, string qchars, vector<int>& qIndices)
    {
        int N = s.length();
        a = s;
        
        Node dummy('a');
        tree.resize(4*N+5, dummy);
        
        // build segment tree
        // 0 indexing segment tree
        buildST(0, 0, N-1);
        
        // for(int i = 0; i < tree.size(); i++)
        //     cout << tree[i].max << endl;
        
        vector<int> ans;
        
        for(int i = 0; i < qIndices.size(); i++)
        {
            int pos = qIndices[i];
            char ch = qchars[i];
            update(0, 0, N-1, pos, ch);
            
            // root of segment tree contains max count
            ans.push_back(tree[0].max);
        }        
        return ans;
    }
};
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