<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>

typedef pair<int,int> Range;
#define LAST(v) (*prev(v.end()))

class Solution {
private:
  set<Range> s;
  map<int, int> lengthCnt;

  int length(int l, int r) {
    return r - l + 1;
  }

  void addRange(int l, int r) {
    s.insert({ l, r });
    ++lengthCnt[length(l, r)];
  }

  void removeRange(set<Range>::iterator it) {
    int x = length(it->first, it->second);
    --lengthCnt[x];
    if (lengthCnt[x] == 0) lengthCnt.erase(x);
    s.erase(it);
  }

  set<Range>::iterator findOverlappingRange(int pos) {
    auto it = s.lower_bound({ pos, pos });
    if (it != s.begin() && prev(it)->second >= pos) --it;
    assert(it->first <= pos && pos <= it->second);
    return it;
  }

  void tryCombineWithNext(const string& str, set<Range>::iterator it) {
    assert(it != s.end());

    int l = it->first, r = next(it)->second;
    // cout << "testing " << l << " " << r << endl;
    if (str[l] != str[r]) return;
    // cout << "combining\n";

    removeRange(next(it));
    removeRange(it);
    addRange(l, r);
  }

  void query(string& str, int pos, char ch) {
    str[pos] = ch;
    // cout << str << endl;

    auto it = findOverlappingRange(pos);
    auto [l, r] = *it;
    removeRange(it);
    addRange(pos, pos);
    if (l < pos) addRange(l, pos-1);
    if (pos < r) addRange(pos+1, r);

    // print();

    it = findOverlappingRange(pos);
    if (it != s.begin()) tryCombineWithNext(str, prev(it));

    it = findOverlappingRange(pos);
    if (next(it) != s.end()) tryCombineWithNext(str, it);

    // print();
  }

  void print() {
    cout << "s: ";
    for (auto r : s) cout << r.first << " " << r.second << ", ";
    cout << endl;
  }

public:
  vector<int> longestRepeating(string str, string queryCharacters, vector<int>& queryIndices) {
    for (int n = str.length(), i = 0; i < n; ++i) {
      int j = i;
      while (j+1 < n && str[i] == str[j+1]) ++j;
      addRange(i, j);
      i = j;
    }

    print();

    vector<int> res;
    for (int n = queryCharacters.length(), i = 0; i < n; ++i) {
      query(str, queryIndices[i], queryCharacters[i]);
      res.push_back(LAST(lengthCnt).first);
    }
    return res;
  }
};
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