<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:
    vector<int> longestRepeating(string s, string queryCharacters, vector<int>& queryIndices) {
        int k = queryCharacters.size();
        
        map<int, int> mp;
        multiset<int> sizes;
        vector<int> res;

        auto insert = [&](int idx, int sz) {
            sizes.insert(sz);
            return mp.insert({idx, sz}).first;
        };

        auto replace = [&](auto p, int new_sz) {
            sizes.erase(sizes.find(p->second));
            sizes.insert(new_sz);
            p->second = new_sz;
        };

        auto erase = [&](auto p) {
            sizes.erase(sizes.find(p->second));
            mp.erase(p);
        };

        auto p = end(mp);
        for(int i = 0; i < s.size(); ++i) {
            if(i == 0 || s[i - 1] != s[i])
                p = mp.insert({i, 1}).first;
            else
                ++p->second;
        }

        for(const auto [i, size] : mp)
            sizes.insert(size);
        
        for(int i = 0; i < k; ++i) {
            char ch = queryCharacters[i];
            int idx = queryIndices[i];

            if(ch != s[idx]) {
                s[idx] = ch;
                auto p = mp.lower_bound(idx);
                if(idx > 0 && (p == mp.end() || p->first > idx)) {
                    auto lastSize = idx - prev(p)->first;
                    p = insert(idx, prev(p)->second - lastSize);
                    replace(prev(p), lastSize);
                }

                if(p->second > 1) {
                    insert(idx + 1, p->second - 1);
                    replace(p, 1);
                }
                
                if(idx < s.size() - 1 && ch == s[idx + 1]) {
                    replace(p, next(p)->second + 1);
                    erase(next(p));
                }

                if(idx > 0 && s[idx - 1] == ch) {
                    replace(prev(p), prev(p)->second + p->second);
                    erase(p);
                }
            }

            res.push_back(*(sizes.rbegin()));
        }

        return res;
    }
};
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