<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) {
        set<int> count;
        map<int, int> dict;
        int last(0);
        count.insert(-1);
        count.insert(0);
        for (int i = 1; i < s.size(); ++i){
            if (s[i] != s[i - 1]) {
                count.insert(i);
                dict[i - last]++;
                last = i;    
            }
        }
        count.insert(s.size());
        dict[s.size() - last]++;
        vector<int> ans(queryCharacters.size(), 0);

        for (int i = 0; i < queryCharacters.size(); ++i){
            // for (auto &a: dict) cout << a.first << ":" << a.second << " ";
            // cout << endl;
            // for (auto &a: count) cout << a << " ";
            // cout << endl;
            if (queryCharacters[i] == s[queryIndices[i]]) ans[i] = dict.rbegin()->first;
            else{
                auto it2 = count.upper_bound(queryIndices[i]);
                auto it1 = prev(it2);
                dict[*it2 - *it1]--;
                if (dict[*it2 - *it1] == 0) dict.erase(dict.find(*it2 - *it1));
                if (queryIndices[i] - *it1 != 0) dict[queryIndices[i] - *it1]++;
                dict[1]++;
                if (*it2 - (queryIndices[i] + 1) != 0) dict[*it2 - queryIndices[i] - 1]++;
                count.insert(queryIndices[i]);
                count.insert(queryIndices[i] + 1);
                s[queryIndices[i]] = queryCharacters[i];
                it2 = count.lower_bound(queryIndices[i]);
                it1 = prev(it2);
                if (*it1 == -1){
                    it1 = next(it1);
                    it2 = next(it2);
                }else{
                    if (s[*it1] == s[*it2]){
                        dict[*it2 - *it1]--;
                        dict[1]--;
                        if (dict[*it2 - *it1] == 0) dict.erase(dict.find(*it2 - *it1));
                        if (dict[1] == 0) dict.erase(dict.find(1));
                        count.erase(it2);
                        it2 = count.upper_bound(queryIndices[i]);
                        dict[*it2 - *it1]++;
                    }else{
                        it1 = next(it1);
                        it2 = next(it2);
                    }
                }

                if (*it2 != s.size()){
                    if (s[*it1] == s[*it2]){
                        dict[*it2 - *it1]--;
                        if (dict[*it2 - *it1] == 0) dict.erase(dict.find(*it2 - *it1));
                        dict[*next(it2) - *it2]--;
                        if (dict[*next(it2) - *it2] == 0) dict.erase(dict.find(*next(it2) - *it2));
                        count.erase(it2);
                        it2 = count.upper_bound(queryIndices[i] + 1);
                        dict[*it2 - *it1]++;
                    }
                }
            }
            ans[i] = dict.rbegin()->first;
        }
        return ans;
    }
};
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