<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 SegNode {
public:
    int range_l;
    int range_r;
    int l_len;
    int r_len;
    int max_len;
    SegNode* left;
    SegNode* right;
    SegNode(int l, int r) {
        range_l = l;
        range_r = r;
    }
};

class SegTree {
private:
    SegNode* root;
    string s;
    SegNode* build(string& s, int l, int r) {
        SegNode* node = new SegNode(l, r);
        if (l == r) {
            node->l_len = 1;
            node->r_len = 1;
            node->max_len = 1;
            return node;
        }
        int mid = (l + r) / 2;
        node->left = build(s, l, mid);
        node->right = build(s, mid + 1, r);
        if (s[mid] == s[mid + 1]) {
            node->l_len = node->left->l_len;
            if (node->left->l_len == (mid - l + 1)) {
                node->l_len = node->left->l_len + node->right->l_len;
            }
            node->r_len = node->right->r_len;
            if (node->right->r_len == (r - mid)) {
                node->r_len = node->right->r_len + node->left->r_len;
            }
            node->max_len = max(node->left->max_len, node->right->max_len);
            node->max_len = max(node->max_len, node->left->r_len + node->right->l_len);
        } else {
            node->l_len = node->left->l_len;
            node->r_len = node->right->r_len;
            node->max_len = max(node->left->max_len, node->right->max_len);
        }
        return node;
    }

    void update(SegNode* node, int idx, char c) {
        if (node->range_l == idx and node->range_r == idx) {
            s[idx] = c;
            return;
        }
        int mid = node->left->range_r;
        if (idx <= mid) {
            update(node->left, idx, c);
        } else {
            update(node->right, idx, c);
        }
        if (s[mid] == s[mid + 1]) {
            node->l_len = node->left->l_len;
            if (node->left->l_len == (mid - node->range_l + 1)) {
                node->l_len = node->left->l_len + node->right->l_len;
            }
            node->r_len = node->right->r_len;
            if (node->right->r_len == (node->range_r - mid)) {
                node->r_len = node->right->r_len + node->left->r_len;
            }
            node->max_len = max(node->left->max_len, node->right->max_len);
            node->max_len = max(node->max_len, node->left->r_len + node->right->l_len);
        } else {
            node->l_len = node->left->l_len;
            node->r_len = node->right->r_len;
            node->max_len = max(node->left->max_len, node->right->max_len);
        }
        return;
    }
public:
    SegTree(string& _s) {
        s = _s;
        root = build(s, 0, s.length() - 1);
    }

    void update_idx(int idx, char c) {
        update(root, idx, c);
    }
    int get_max() {
        return root->max_len;
    }
};


class Solution {
public:
    vector<int> longestRepeating(string s, string queryCharacters, vector<int>& queryIndices) {
        SegTree seg_tree(s);
        int N = queryIndices.size();
        vector<int> ans(N, 0);
        for (int i = 0; i < N; i++) {
            seg_tree.update_idx(queryIndices[i], queryCharacters[i]);
            ans[i] = seg_tree.get_max();
        }
        return ans;
    }
};

// Q: length of queries, N: length of s
// time: O(QlogN)
// space: O(N + Q)

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