<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>

struct Node{
public:
    int sz;
    int pref_l, suff_l, best_l;
    char pref_c, suff_c;
    Node(int sz=0, int pref_l=0, int suff_l=0, int best_l=0, char pref_c='#', char suff_c='#'): sz(sz), pref_l(pref_l), suff_l(suff_l), best_l(best_l), pref_c(pref_c), suff_c(suff_c){}
};
class SGTree{
public:
    vector<Node*> segarr;
    SGTree(int n=0){
        this->segarr.resize(4*n + 1, NULL);
    }

    inline void updateNode(Node* root, int sz=0, int pref_l=0, int suff_l=0, int best_l=0, char pref_c='#', char suff_c='#'){
        root->sz = sz;
        root->pref_l = pref_l;
        root->pref_c = pref_c;
        root->suff_l = suff_l;
        root->suff_c = suff_c;
        root->best_l = best_l;
    }
    
    inline Node* merge(Node* n1, Node* n2){
        Node* res = new Node();
    
        res->sz = n1->sz + n2->sz;
        
        // merge prefix
        if(n1->pref_l!=n1->sz || n1->pref_c!=n2->pref_c){
            // merging does not occur when length is less or character mismatch
            res->pref_l = n1->pref_l;
            res->pref_c = n1->pref_c;
        }else{
            // merging will occur as length is full and character also matches
            res->pref_l = n1->pref_l+n2->pref_l;
            res->pref_c = n1->pref_c;
        }
        
        // merge suffix
        if(n2->suff_l!=n2->sz || n2->suff_c!=n1->suff_c){
            // merging does not occur when length is less or character mismatch
            res->suff_l = n2->suff_l;
            res->suff_c = n2->suff_c;
        }else{
            // merging will occur as length is full and character also matches
            res->suff_l = n2->suff_l+n1->suff_l;
            res->suff_c = n2->suff_c;
        }
        
        // get the best string
        res->best_l = max({res->pref_l, res->suff_l, n1->best_l, n2->best_l});
        if(n1->suff_c==n2->pref_c){
            res->best_l = max({res->best_l, n1->suff_l+n2->pref_l});
        }
        
        return res;
    }
    
    void build(int index, int low, int high, string& s){
        if(low == high){
            // updateNode(this->segarr[index], 1, 1, 1, 1, s[low], s[low]);
            this->segarr[index] = new Node(1, 1, 1, 1, s[low], s[low]);
            return;
        }
        int mid = low + ((high - low) >> 1);
        build(2*index+1, low, mid, s);
        build(2*index+2, mid+1, high, s);
        this->segarr[index] = merge(this->segarr[2*index+1], this->segarr[2*index+2]);
    }

    void update(int index, int low, int high, int ind, char ch){
        if(low==high){
            // update
            updateNode(this->segarr[index], 1, 1, 1, 1, ch, ch);
            // this->segarr[index] = new Node(1, 1, 1, 1, ch, ch);
            return;
        }
        int mid = low + ((high - low) >> 1);
        if(ind <= mid){
            update(2*index+1, low, mid, ind, ch);
        }else{
            update(2*index+2, mid+1, high, ind, ch);
        }
        this->segarr[index] = merge(this->segarr[2*index+1], this->segarr[2*index+2]);
    }
};
class Solution {
public:
    vector<int> longestRepeating(string s, string queryCharacters, vector<int>& queryIndices) {
        int n = s.length();
        int k = queryIndices.size();
        SGTree *sgtree = new SGTree(n);
        sgtree->build(0, 0, n-1, s);
        // for(int i=0; i<n; i++) sgtree->update(0, 0, n-1, i, s[i]);
        vector<int> ans;
        for(int i=0; i<k; i++){
            sgtree->update(0, 0, n-1, queryIndices[i], queryCharacters[i]);
            ans.push_back(sgtree->segarr[0]->best_l);
        }
        return ans;
    }
};
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