<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:
    struct Node {
        int lc;
        char ls;
        int rc;
        char rs;
        int mx;
        int sz;
    };

    Node mergeNode(Node a, Node b) {
        int mx = max(a.mx, b.mx);
        if (a.rs == b.ls) {
            mx = max(mx, a.rc + b.lc);
        }

        int left_len = a.lc;
        if (a.lc == a.sz && b.ls == a.ls) {
            left_len += b.lc;
        }

        int right_len = b.rc;
        if (b.rc == b.sz && a.rs == b.rs) {
            right_len += a.rc;
        }
        return Node{
            left_len,
            a.ls,
            right_len,
            b.rs,
            mx,
            a.sz + b.sz,
        };
    }

    Node createNode(char c) {
        return Node{
            1,
            c,
            1,
            c,
            1,
            1,
        };
    }

    vector<Node> tree;
    string sr;

    void init(int i, int l, int r) {
        if (l == r) {
            tree[i] = createNode(sr[l]);
            return;
        }

        int m = (l + r) / 2;
        init(2 * i, l, m);
        init(2 * i + 1, m + 1, r);
        tree[i] = mergeNode(tree[2 * i], tree[2 * i + 1]);
    }

    void upd(int i, int l, int r, int pos, char c) {
        if (l == r) {
            tree[i] = createNode(c);
            return;
        }

        int m = (l + r) / 2;
        if (pos <= m) {
            upd(2 * i, l, m, pos, c);
        } else {
            upd(2 * i + 1, m + 1, r, pos, c);
        }
        
        tree[i] = mergeNode(tree[2 * i], tree[2 * i + 1]);
    }

    vector<int> longestRepeating(string s, string queryCharacters, vector<int>& queryIndices) {
        int n = s.size();
        tree.resize(4 * n);
        sr = s;

        init(1, 0, n - 1);

        int k = queryCharacters.size();
        vector<int> res(k);

        for (int i = 0; i < k; ++i) {
            upd(1, 0, n - 1, queryIndices[i], queryCharacters[i]);

            res[i] = tree[1].mx;
        }
        return res;
    }
};
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