<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 Node {
public:
    int left, right, size, maxLen, leftMax, rightMax;
    char leftChar, rightChar;

    Node(): 
        left(-1), right(-1), size(0), maxLen(0), leftChar('$'), rightChar('$'), leftMax(0), rightMax(0)
        {} 
};

class SegmentTree {
private:
    void merge(int node) {
        int leftNode = 2 * node + 1;
        int rightNode = 2 * node + 2;

        tree[node] = merge(leftNode, rightNode);
    }

    Node merge(int lNode, int rNode) {
        Node res;

        Node left = tree[lNode];
        Node right = tree[rNode];

        // Correct size calculation
        res.size = left.size + right.size;
        res.left = left.left;
        res.right = right.right;
        res.leftChar = left.leftChar;
        res.rightChar = right.rightChar;

        res.maxLen = max(left.maxLen, right.maxLen);
        res.leftMax = left.leftMax;
        res.rightMax = right.rightMax;

        // Properly merge leftMax and rightMax
        if (left.rightChar == right.leftChar) {
            if (left.leftMax == left.size) {
                res.leftMax += right.leftMax;
            }
            if (right.rightMax == right.size) {
                res.rightMax += left.rightMax;
            }
            res.maxLen = max(res.maxLen, left.rightMax + right.leftMax);
        }

        return res;
    }

public:
    vector<Node> tree;
    string str;

    SegmentTree(string &s): str(s) {
        int n = s.length();
        tree.resize(n << 2);
        build(0, 0, n-1);
    }

    void build(int node, int l, int r) {
        tree[node].left = l;
        tree[node].right = r;

        if (l == r) {
            tree[node].leftMax = tree[node].rightMax = tree[node].maxLen = tree[node].size = 1;
            tree[node].leftChar = tree[node].rightChar = str[l];
            return;
        }

        int mid = (l + r) >> 1;
        build(node * 2 + 1, l, mid);
        build(node * 2 + 2, mid + 1, r);

        merge(node);
    }

    void update(int node, int ind, char ch, int l, int r) {
        if (l == r) {
            tree[node].leftChar = tree[node].rightChar = ch;
            tree[node].leftMax = tree[node].rightMax = tree[node].maxLen = 1;
            return;
        }

        int mid = (l + r) >> 1;

        if (mid >= ind) {
            update(node * 2 + 1, ind, ch, l, mid);
        } else {
            update(node * 2 + 2, ind, ch, mid + 1, r);
        }

        merge(node);
    }
};

class Solution {
public:
    vector<int> longestRepeating(string s, string queryChars, vector<int>& queryIndices) {
        SegmentTree segTree(s);
        int numQueries = queryChars.size();
        vector<int> result(numQueries);

        for (int i = 0; i < numQueries; ++i) {
            int pos = queryIndices[i];
            segTree.update(0, pos, queryChars[i], 0, s.length() - 1);
            result[i] = segTree.tree[0].maxLen;
        }

        return result;
    }
};

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