<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) {
        vector<set<pair<int, int>>> pts(26);
        multiset<int> len;
        vector<int> res;

        for(int i = 0; i < s.size(); ++i) {
            char x = s[i];
            int left = i, right = i;

            while(i + 1 < s.size() and s[i + 1] == x) {
                right = i + 1;
                ++i;
            }

            len.insert(-(right - left + 1));

            pts[x - 'a'].insert({left, right});
        }

        for(int i = 0; i < queryIndices.size(); ++i) {
            int index = queryIndices[i];
            char x = s[index];
            char newX = queryCharacters[i];

            if(newX == x) {
                res.push_back(-(*len.begin()));
                continue;
            }

            auto it = pts[x - 'a'].lower_bound({index, 0});

            if(it != begin(pts[x - 'a']) and index <= prev(it)->second) it = prev(it);

            if(it->first == it->second) {
                auto lit = len.find(-(it->second - it->first + 1));
                len.erase(lit);
                pts[x - 'a'].erase(it);
            } else {
                if(it->first == index) {
                    auto lit = len.find(-(it->second - it->first + 1));
                    len.erase(lit);
                    len.insert(-(it->second - (index + 1) + 1));
                    pts[x - 'a'].insert({index + 1, it->second});
                    pts[x - 'a'].erase(it);
                } else if(it->second == index) {
                    auto lit = len.find(-(it->second - it->first + 1));
                    len.erase(lit);
                    len.insert(-((index - 1) - it->first + 1));
                    pts[x - 'a'].insert({it->first, index - 1});
                    pts[x - 'a'].erase(it);
                } else {
                    auto lit = len.find(-(it->second - it->first + 1));
                    len.erase(lit);
                    pts[x - 'a'].insert({it->first, index - 1});
                    pts[x - 'a'].insert({index + 1, it->second});
                    len.insert(-(index - 1 - it->first + 1));
                    len.insert(-(it->second - (index + 1) + 1));
                    pts[x - 'a'].erase(it);
                }
            }

            auto it2 = pts[newX - 'a'].lower_bound({index, 0});
            int left = index, right = index;

            if(it2 != begin(pts[newX - 'a']) and prev(it2)->second + 1 == index) {
                left = prev(it2)->first;
                auto lit = len.find(-(prev(it2)->second - prev(it2)->first + 1));
                len.erase(lit);
                pts[newX - 'a'].erase(prev(it2));
            }

            if(it2 != end(pts[newX - 'a']) and it2->first == index + 1) {
                right = it2->second;
                auto lit = len.find(-(it2->second - it2->first + 1));
                len.erase(lit);
                pts[newX - 'a'].erase(it2);
            }

            len.insert(-(right - left + 1));
            pts[newX - 'a'].insert({left, right});
            s[index] = newX;
            res.push_back(-(*len.begin()));
        }

        return res;
    }
};
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