<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 ds{
        int longSbstr = 0;
        char leftChar;
        char rightChar;
        int prefixLen = 1;
        int suffixLen = 1;
        int totalLen = 1;
    };

    vector<ds> buildST(string s, int& n) {
        
        vector<ds> ST (4*n);
        int i = 1;
        while (true) {
            if (i >= n)
                break;
            i *= 2;
        }
        int newN = i;  

        for (int j = 0; j < n; j++){
            ST[newN - 1 + j].longSbstr = 1;
            ST[newN - 1 + j].leftChar = s[j];
            ST[newN - 1 + j].rightChar = s[j];
        }
        i = newN - 2;
        n = newN;
        while (i > -1) {

            ds leftNode = ST[2*i+1];
            ds rightNode = ST[2*i+2];

            if (leftNode.longSbstr == 0)
                ST[i] = ST[i];
            else if (rightNode.longSbstr == 0)
                ST[i] = leftNode;
            else{
                ST[i].leftChar = leftNode.leftChar;
                ST[i].rightChar = rightNode.rightChar;
                ST[i].totalLen = leftNode.totalLen + rightNode.totalLen;
                ST[i].longSbstr = max(leftNode.longSbstr, rightNode.longSbstr);
                ST[i].prefixLen = leftNode.prefixLen;
                ST[i].suffixLen = rightNode.suffixLen;
                if (leftNode.rightChar == rightNode.leftChar) { // There is a new repeating character sequence formed in the middle
                    ST[i].longSbstr = max(ST[i].longSbstr, leftNode.suffixLen + rightNode.prefixLen);
                    if (leftNode.prefixLen == leftNode.totalLen) // Entire left substring has repeating characters
                        ST[i].prefixLen = leftNode.totalLen + rightNode.prefixLen;
                    if (rightNode.prefixLen == rightNode.totalLen)
                        ST[i].suffixLen = leftNode.suffixLen + rightNode.totalLen;
                }                
            }
            i--;
        }
        return ST;
    }
    void updateST(vector<ds>& ST, int i, char c, int n) {
        i += n - 1;
        ST[i].leftChar = c;
        ST[i].rightChar = c;
        while (i > 0) {
            i = (i - 1) / 2;
            ds leftNode = ST[2*i+1];
            ds rightNode = ST[2*i+2];
            if (leftNode.longSbstr == 0)
                ST[i] = ST[i];
            else if (rightNode.longSbstr == 0)
                ST[i] = leftNode;
            else{
                ST[i].leftChar = leftNode.leftChar;
                ST[i].rightChar = rightNode.rightChar;
                ST[i].totalLen = leftNode.totalLen + rightNode.totalLen;
                ST[i].longSbstr = max(leftNode.longSbstr, rightNode.longSbstr);
                ST[i].prefixLen = leftNode.prefixLen;
                ST[i].suffixLen = rightNode.suffixLen;
                if (leftNode.rightChar == rightNode.leftChar) { // There is a new repeating character sequence formed in the middle
                    ST[i].longSbstr = max(ST[i].longSbstr, leftNode.suffixLen + rightNode.prefixLen);
                    if (leftNode.prefixLen == leftNode.totalLen) // Entire left substring has repeating characters
                        ST[i].prefixLen = leftNode.totalLen + rightNode.prefixLen;
                    if (rightNode.prefixLen == rightNode.totalLen)
                        ST[i].suffixLen = leftNode.suffixLen + rightNode.totalLen;
                }
            }
        }
    }
    vector<int> longestRepeating(string s, string queryCharacters, vector<int>& queryIndices) {
        int n = s.size();
        vector<ds> ST = buildST(s, n);
        vector<int> res;
        for (int i = 0; i < queryIndices.size(); i++) {
            updateST(ST, queryIndices[i], queryCharacters[i], n);
            res.push_back(ST[0].longSbstr);
        }
        return res;
    }
};
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