<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 Node {
public:
    int l, r;
    int lmx, rmx, mx;

    Node(int l, int r)
        : l(l)
        , r(r)
        , lmx(1)
        , rmx(1)
        , mx(1) {}
};

class SegmentTree {
private:
    string s;
    vector<Node*> tr;

    void build(int u, int l, int r) {
        tr[u] = new Node(l, r);
        if (l == r) {
            return;
        }
        int mid = (l + r) >> 1;
        build(u << 1, l, mid);
        build(u << 1 | 1, mid + 1, r);
        pushup(u);
    }

    void pushup(int u) {
        Node* root = tr[u];
        Node* left = tr[u << 1];
        Node* right = tr[u << 1 | 1];
        root->mx = max(left->mx, right->mx);
        root->lmx = left->lmx;
        root->rmx = right->rmx;
        int a = left->r - left->l + 1;
        int b = right->r - right->l + 1;
        if (s[left->r - 1] == s[right->l - 1]) {
            if (left->lmx == a) {
                root->lmx += right->lmx;
            }
            if (right->rmx == b) {
                root->rmx += left->rmx;
            }
            root->mx = max(root->mx, left->rmx + right->lmx);
        }
    }

public:
    SegmentTree(const string& s)
        : s(s) {
        int n = s.size();
        tr.resize(n * 4);
        build(1, 1, n);
    }

    void modify(int u, int x, char v) {
        if (tr[u]->l == x && tr[u]->r == x) {
            s[x - 1] = v;
            return;
        }
        int mid = (tr[u]->l + tr[u]->r) >> 1;
        if (x <= mid) {
            modify(u << 1, x, v);
        } else {
            modify(u << 1 | 1, x, v);
        }
        pushup(u);
    }

    int query(int u, int l, int r) {
        if (tr[u]->l >= l && tr[u]->r <= r) {
            return tr[u]->mx;
        }
        int mid = (tr[u]->l + tr[u]->r) >> 1;
        int ans = 0;
        if (r <= mid) {
            ans = query(u << 1, l, r);
        } else if (l > mid) {
            ans = max(ans, query(u << 1 | 1, l, r));
        }
        return ans;
    }
};

class Solution {
public:
    vector<int> longestRepeating(string s, string queryCharacters, vector<int>& queryIndices) {
        SegmentTree tree(s);
        int k = queryIndices.size();
        vector<int> ans(k);
        int n = s.size();
        for (int i = 0; i < k; ++i) {
            int x = queryIndices[i] + 1;
            char v = queryCharacters[i];
            tree.modify(1, x, v);
            ans[i] = tree.query(1, 1, n);
        }
        return ans;
    }
};
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