<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>

struct Node
{
    int maxLength, leftLength, rightLength;
    char leftChar, rightChar;
    Node() : leftChar(0), rightChar(0), maxLength(1), rightLength(1), leftLength(1) {}
};

class SegmentTree
{
    public:
    vector<Node> root;
    int n;
    SegmentTree(const string& s)
    {
        this->n = s.size();
        root.resize(n * 4);
        build(0, 0, n - 1, s);
    }

    void build(int i, int left, int right, const string& s)
    {
        if(left == right)
        {
            root[i].leftChar = root[i].rightChar = s[left];
            return;
        }
        int middle = left + (right - left) / 2;
        build(i * 2 + 1, left, middle, s);
        build(i * 2 + 2, middle + 1, right, s);
        merge(i, left, middle, right);
    }

    void update(int index, char ch)
    {
        update(0, 0, n - 1, index, ch);
    }

    void update(int i, int left, int right, int index, char ch)
    {
        if(left == right)
        {
            root[i].leftChar = root[i].rightChar = ch;
            return;
        }
        int middle = left + (right - left) / 2;
        if(index <= middle)
        {
            update(i * 2 + 1, left, middle, index, ch);
        }
        else
        {
            update(i * 2 + 2, middle + 1, right, index, ch);
        }
        merge(i, left, middle, right);
    }
    void merge(int i, int left, int middle, int right)
    {
        const Node& leftChild = root[i * 2 + 1];
        const Node& rightChild = root[i * 2 + 2];
        root[i].leftChar = leftChild.leftChar;
        root[i].rightChar = rightChild.rightChar;
        root[i].leftLength = leftChild.leftLength;
        if(leftChild.leftLength == middle - left + 1 && leftChild.rightChar == rightChild.leftChar)
        {
            root[i].leftLength += rightChild.leftLength;
        }
        root[i].rightLength = rightChild.rightLength;
        if(rightChild.rightLength == right - middle && rightChild.leftChar == leftChild.rightChar)
        {
            root[i].rightLength += leftChild.rightLength;
        }
        root[i].maxLength = max(leftChild.maxLength, rightChild.maxLength);
        if(leftChild.rightChar == rightChild.leftChar)
        {
            root[i].maxLength = max(root[i].maxLength, leftChild.rightLength + rightChild.leftLength);
        }
    }

    int get()
    {
        return root[0].maxLength;
    }
};

class Solution {
public:
    vector<int> longestRepeating(string s, string qc, vector<int>& qi) {
        // SOLUTION OF TIN LE
        ios_base::sync_with_stdio(false);
        cin.tie(nullptr);
        cout.tie(nullptr);
        vector<int> res;
        SegmentTree root(s);
        for(int i = 0; i < qc.size(); i++)
        {
            int index = qi[i];
            char ch = qc[i];
            root.update(index, ch);
            res.push_back(root.get());
        }
        return res;
    }
};
101000