<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 tuple<int,int,char> Range;
#define FIRST(v) v[0]
#define LAST(v) v[(int) v.size() - 1]

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

  void print() {
    for (auto [l,r,c] : s) cout << l << " " << r << " " << c << ", ";
    cout << endl;
  }

  int length(const Range& r) {
    return get<1>(r) - get<0>(r) + 1;
  }

  set<Range>::iterator findIterator(int pos) {
    auto it = s.lower_bound({ pos, pos, 0 });
    if (it == s.end()) --it;
    if (get<0>(*it) > pos) --it;
    assert(get<0>(*it) <= pos && pos <= get<1>(*it));
    return it;
  }

  set<Range>::iterator addRange(const Range& r) {
    ++cnt[length(r)];
    return s.insert(r).first;
  }

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

  void add(int pos, char ch) {
    auto it = findIterator(pos);
    auto [l, r, c] = *it;
    removeRange(it);

    if (l < pos) addRange(make_tuple(l, pos-1, c));
    if (pos < r) addRange(make_tuple(pos+1, r, c));
    auto z = addRange({ pos, pos, ch });

    if (z != s.begin() && get<2>(*prev(z)) == ch) --z;

    vector<set<Range>::iterator> p;
    while (z != s.end() && get<2>(*z) == ch) {
      p.push_back(z);
      ++z;
    }

    assert(!p.empty());
    l = min( pos, get<0>(*FIRST(p)) );
    r = max( pos, get<1>(*LAST(p)) );

    for (auto range : p) removeRange(range);
    addRange(make_tuple(l, r, ch));
  }

public:
  vector<int> longestRepeating(string str, string queryCharacters, vector<int>& queryIndices) {
    int n = str.length();

    for (int i = 0; i < n; ++i) {
      int j = i;
      while (j+1 < n && str[j+1] == str[i]) ++j;
      addRange(make_tuple(i, j, str[i]));
      i = j;
    }

    vector<int> res;
    for (int i = 0; i < (int) queryCharacters.size(); ++i) {
      add(queryIndices[i], queryCharacters[i]);
      res.push_back(prev(cnt.end())->first);
      // cout << prev(cnt.end())->first << " -> " << prev(cnt.end())->second << endl;
    }

    return res;
  }
};
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