<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:
    void build(int ind, int low, int high, vector<vector<int>>& st, string& s) {
        if(low == high) {
            st[ind] = {s[low], 1, s[low], 1, 1};
            return;
        }
        int mid = (high + low) >> 1;
        int left = 2 * ind + 1;
        int right = 2 * ind + 2;
        build(left, low, mid, st, s);
        build(right, mid + 1, high, st, s);

        st[ind][0] = st[left][0];
        st[ind][2] = st[right][2];
        int temp = 0;
        if(st[left][2] == st[right][0]) {
            temp = st[right][1] + st[left][3];
        }
        st[ind][1] = (st[left][1] + st[left][3] == 2 * (mid - low + 1) && temp > 0) ? temp : st[left][1];
        st[ind][3] = (st[right][1] + st[right][3] == 2 * (high - mid) && temp > 0) ? temp : st[right][3];
        st[ind][4] = max(temp, max(st[left][4], st[right][4]));
    }

    void update(int ind, int low, int high, vector<vector<int>>& st, int& target, char& ch) {
        if(low == high) {
            st[ind] = {ch, 1, ch, 1, 1};
            return;
        }
        int mid = (high + low) >> 1;
        int left = 2 * ind + 1;
        int right = 2 * ind + 2;
        if(low <= target && mid >= target) update(left, low, mid, st, target, ch);
        else update(right, mid + 1, high, st, target, ch);

        st[ind][0] = st[left][0];
        st[ind][2] = st[right][2];
        int temp = 0;
        // cout << "At index " << ind << endl;
        // cout << "left child: " << char(st[left][0]) << " " << st[left][1] << " " << char(st[left][2]) << " " << st[left][3] << " " << st[left][4] << endl;
        // cout << "right child: " << char(st[right][0]) << " " << st[right][1] << " " << char(st[right][2]) << " " << st[right][3] << " " << st[right][4] << endl << endl;
        if(st[left][2] == st[right][0]) {
            temp = st[right][1] + st[left][3];
        }
        st[ind][1] = (st[left][1] + st[left][3] == 2 * (mid - low + 1) && temp > 0) ? temp : st[left][1];
        st[ind][3] = (st[right][1] + st[right][3] == 2 * (high - mid) && temp > 0) ? temp : st[right][3];
        st[ind][4] = max(temp, max(st[left][4], st[right][4]));
    }
    vector<int> longestRepeating(string s, string qc, vector<int>& qi) {
        int n = s.size();
        vector<vector<int>> st(5*n, vector<int> (5, 0));
        build(0, 0, n - 1, st, s);

        vector<int> ans;
        for(int i = 0; i < qc.size(); i++) {
            update(0, 0, n - 1, st, qi[i], qc[i]);
            ans.push_back(st[0][4]);
        }

        return ans;
    }
};
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