<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 qc, vector<int>& qi) {
        map<int, int> lcnt;
        set<pair<int, int>> ranges;
        int m = s.length();
        for (int i = 0; i < m; ) {
            int j = i + 1;
            while (j < m && s[j] == s[i]) {
                ++j;
            }
            ++lcnt[j - i];
            ranges.emplace(i, j - 1);
            i = j;
        }
        vector<int> res;
        int k = qi.size();
        for (int i = 0; i < k; ++i) {
            char c = qc[i];
            int idx = qi[i];
            if (c != s[idx]) {
                auto range = *prev(ranges.upper_bound({ idx, INT_MAX }));
                removeRange(lcnt, ranges, range);
                addRange(lcnt, ranges, { idx, idx });
                if (range.first < idx) {
                    addRange(lcnt, ranges, { range.first, idx - 1 });
                }
                if (idx < range.second) {
                    addRange(lcnt, ranges, { idx + 1, range.second });
                }
                if (idx > 0 && s[idx - 1] == c) {
                    removeRange(lcnt, ranges, { idx, idx });
                    auto it = prev(ranges.upper_bound({ idx - 1, INT_MAX }));
                    addRange(lcnt, ranges, { it->first, idx });
                    removeRange(lcnt, ranges, *it);
                }
                if (idx + 1 < m && s[idx + 1] == c) {
                    auto it1 = prev(ranges.upper_bound({ idx, INT_MAX }));
                    auto it2 = prev(ranges.upper_bound({ idx + 1, INT_MAX }));
                    addRange(lcnt, ranges, { it1->first, it2->second });
                    removeRange(lcnt, ranges, *it1);
                    removeRange(lcnt, ranges, *it2);
                }
            }
            s[idx] = c;
            res.push_back(rbegin(lcnt)->first);
        }
        return res;
    }
private:
    void addRange(map<int, int>& lcnt, set<pair<int, int>>& ranges, const pair<int, int>& range) {
        ++lcnt[range.second - range.first + 1];
        ranges.insert(range);
    }

    void removeRange(map<int, int>& lcnt, set<pair<int, int>>& ranges, const pair<int, int>& range) {
        int l = range.second - range.first + 1;
        auto it = lcnt.find(l);
        if (--it->second == 0) {
            lcnt.erase(it);
        }
        ranges.erase(range);
    }
};
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