<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 val;    //length of longest substr of same char
        int ss;     //start of substring
        int se;     //end of substring
        char lc;    //left most char
        char rc;    //right most char
        int rlc;    //right most index of contiguous left most char (i.e. lc) 
        int lrc;    //left most index of contiguous right most char (i.e. rc)
    };

    void merge( vector<Node>& ST, int ci, int l, int m, int r ){
        int leftC = 2*ci + 1;
        int rightC = 2*ci + 2;

        ST[ci].lc = ST[leftC].lc;
        ST[ci].rc = ST[rightC].rc;
        ST[ci].rlc = ST[leftC].rlc;
        ST[ci].lrc = ST[rightC].lrc;
        ST[ci].ss = ST[leftC].ss;
        ST[ci].se = ST[rightC].se;

        int maxLen = 0;
        m = (ST[ci].ss + ST[ci].se) / 2;
        if( ST[leftC].rc == ST[rightC].lc ){
            if( ST[leftC].rlc == ST[leftC].se ) 
                ST[ci].rlc = ST[rightC].rlc;
            if( ST[rightC].lrc == ST[rightC].ss ) 
                ST[ci].lrc = ST[leftC].lrc;
            maxLen = ST[rightC].rlc - ST[leftC].lrc + 1;
        }

        maxLen = max( maxLen, max(ST[leftC].rlc - l + 1, r - ST[rightC].lrc + 1) );
        maxLen = max( maxLen, max(ST[leftC].val, ST[rightC].val) );

        ST[ci].val = maxLen;
        
    }    

    void buildST( vector<Node>& ST, string& s, int ci, int l, int r){
        
        for( int i = 0; i < s.size(); i++ ){
            int currI = ST.size() / 2 + i;
            ST[currI].lc = ST[currI].rc = s[i];
            ST[currI].rlc = ST[currI].lrc = i;
            ST[currI].ss = ST[currI].se = i;
            ST[currI].val = 1;
        }

        long long p = ST.size() / 2 - 1;
        while( p >= 0 ){
            int ss = ST[2*p + 1].ss;
            int se = ST[2*p + 2].se;

            if( ST[2*p + 1].val == 0 && ST[2*p + 2].val == 0 ){
                p--;
                continue;
            }
            else if( ST[2*p + 1].val == 0 && ST[2*p + 2].val > 0 ){
                ST[p] = ST[2*p + 2];
                p--;
                continue;
            }
            else if( ST[2*p + 1].val > 0 && ST[2*p + 2].val == 0 ){
                ST[p] = ST[2*p + 1];
                p--;
                continue;
            }

            merge( ST, p, ss, (ss+se)/2, se);
            p--;
        }

    }

    void applyQuery( vector<Node>& ST, string& s, int ci, char qc ){
        ST[ci].lc = ST[ci].rc = qc;

        int p = ci;
        while( p > 0 ){
            p = (p - 1) / 2;

             if( ST[2*p + 1].val == 0 && ST[2*p + 2].val == 0 ){
                continue;
            }
            else if( ST[2*p + 1].val == 0 && ST[2*p + 2].val > 0 ){
                ST[p] = ST[2*p + 2];
                continue;
            }
            else if( ST[2*p + 1].val > 0 && ST[2*p + 2].val == 0 ){
                ST[p] = ST[2*p + 1];
                continue;
            }

            merge( ST, p, ST[p].ss, (ST[p].ss + ST[p].se) / 2, ST[p].se  );
        }

    }

    vector<int> longestRepeating(string s, string queryCharacters, vector<int>& queryIndices) {
        ios_base::sync_with_stdio(false);
        cin.tie(nullptr);
        cout.tie(nullptr);

        double logVal = log2( s.size() );
        int len = s.size();
        int stlen = 2 * pow(2, ceil(logVal)) - 1;
        vector<struct Node> ST ( stlen, Node(0 , -1, -1, '_', '_', -1, -1) );
        vector<int> v;

        buildST( ST, s, 0, 0, len - 1 );

        for( int i = 0; i < queryIndices.size(); i++ ){
            applyQuery( ST, s, ST.size() / 2 + queryIndices[i], queryCharacters[i] );
            // cout << ST[0].val << endl;
            v.push_back( ST[0].val );
        }   

        return v;
    }
};
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