<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 SegmentTree {
private:
    struct Node {
        int maxLen, prefixLen, suffixLen, totalLen;
        char prefixChar, suffixChar;
        
        Node(int len = 0, char c = '#') : 
            maxLen(len), prefixLen(len), suffixLen(len), totalLen(len),
            prefixChar(c), suffixChar(c) {}
    };
    
    vector<Node> tree;
    int n;
    
    void build(const string& s, int v, int tl, int tr) {
        if (tl == tr) {
            tree[v] = Node(1, s[tl]);
            return;
        }
        
        int tm = (tl + tr) / 2;
        build(s, v*2, tl, tm);
        build(s, v*2+1, tm+1, tr);
        merge(v);
    }
    
    void merge(int v) {
        Node& left = tree[v*2];
        Node& right = tree[v*2+1];
        Node& current = tree[v];
        
        current.totalLen = left.totalLen + right.totalLen;
        current.prefixChar = left.prefixChar;
        current.suffixChar = right.suffixChar;
        
        current.prefixLen = left.prefixLen;
        if (left.prefixLen == left.totalLen && left.suffixChar == right.prefixChar)
            current.prefixLen += right.prefixLen;
        
        current.suffixLen = right.suffixLen;
        if (right.suffixLen == right.totalLen && left.suffixChar == right.prefixChar)
            current.suffixLen += left.suffixLen;
        
        current.maxLen = max({left.maxLen, right.maxLen, 
                              (left.suffixChar == right.prefixChar) ? 
                              left.suffixLen + right.prefixLen : 0});
    }
    
    void update(int v, int tl, int tr, int pos, char newChar) {
        if (tl == tr) {
            tree[v] = Node(1, newChar);
            return;
        }
        
        int tm = (tl + tr) / 2;
        if (pos <= tm)
            update(v*2, tl, tm, pos, newChar);
        else
            update(v*2+1, tm+1, tr, pos, newChar);
        
        merge(v);
    }
    
public:
    SegmentTree(const string& s) : n(s.length()) {
        tree.resize(4 * n);
        if (!s.empty()) {
            build(s, 1, 0, n-1);
        }
    }
    
    void update(int pos, char newChar) {
        if (n > 0) {
            update(1, 0, n-1, pos, newChar);
        }
    }
    
    int getMaxLen() {
        return n > 0 ? tree[1].maxLen : 0;
    }
};

class Solution {
public:
    vector<int> longestRepeating(string s, string queryCharacters, vector<int>& queryIndices) {
        SegmentTree segTree(s);
        int k = queryCharacters.length();
        vector<int> result(k);
        
        for (int i = 0; i < k; ++i) {
            segTree.update(queryIndices[i], queryCharacters[i]);
            result[i] = segTree.getMaxLen();
        }
        
        return result;
    }
};
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