<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>

#include <bits/stdc++.h>

using namespace std;

class Solution {
public:
    class node {
    public:
        char char_l, char_r;
        int ans_l, ans_r, ans, len;
        node() : char_l(0), char_r(0), ans_l(0), ans_r(0), ans(0), len(0) {}
        node(const char& x) : char_l(x), char_r(x), ans_l(1), ans_r(1), ans(1), len(1) {}
        node operator + (const node& other) {
            node ret;
            ret.char_l = this->char_l;
            ret.char_r = other.char_r;
            ret.ans_l = this->ans_l;
            ret.ans_r = other.ans_r;
            ret.len = this->len + other.len;
            ret.ans = max(this->ans, other.ans);
            if (this->char_r == other.char_l) {
                ret.ans = max(ret.ans, this->ans_r + other.ans_l);
            }
            if (other.ans_r ==  other.len) {
                if (other.char_r == this->char_r) {
                    ret.ans_r += this->ans_r;
                }
            }
            if (this->ans_l ==  this->len) {
                if (this->char_l == other.char_l) {
                    ret.ans_l += other.ans_l;
                }
            }
            return ret;
        }
    };
    
    vector < node > it;

    void build(const int& n) {
        it.resize(4 * n + 1);
    }

    void update(int id, int l, int r, int pos, char x) {
        if (pos < l || pos > r) return;
        if (l == r) {
            it[id] = node(x);
            return;
        }
        int m = l + r >> 1;
        update(id << 1, l, m, pos, x);
        update(id << 1 | 1, m + 1, r, pos, x);
        it[id] = it[id << 1] + it[id << 1 | 1];
    }

    vector<int> longestRepeating(string s, string queryCharacters, vector<int>& queryIndices) {
        int n = s.size();
        build(n);
        for (int i = 0; i < n; i ++) {
            update(1, 0, n - 1, i, s[i]);
        }

        vector < int > ans;

        for (int t = 0; t < queryCharacters.size(); t ++) {
            char x = queryCharacters[t];
            int id = queryIndices[t];
            update(1, 0, n - 1, id, x);
            ans.emplace_back(it[1].ans);
        }
        
        return ans;
    }
};
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