<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
{

public:
    string s = "";
    vector<vector<int>> seg;
    SegmentTree(int n, string s1)
    {
        s = s1;
        seg.resize(4 * n, {0, 0, 0, 0});
    }
    void build(int l, int r, int idx)
    {
        if (l == r)
        {
            // seg[idx] = {len,left,right};
            seg[idx] = {1, 1, 1};
            return;
        }
        int mid = (l + r) / 2;
        build(l, mid, 2 * idx + 1);
        build(mid + 1, r, 2 * idx + 2);
        int len1 = seg[2 * idx + 1][2];
        int len2 = seg[2 * idx + 2][1];
        int mxleft = seg[2 * idx + 1][0];
        int mxright = seg[2 * idx + 2][0];
        // if(idx==1){
        //     cout<<len1<<" "<<len2<<" "<<mxleft<<" "<<mxright<<endl;
        // }
       if ((s[mid] == s[mid + 1]))
        {

            if ((len1 + len2 >= max(mxleft, mxright)))
            {
                seg[idx][0] = len1 + len2;
                seg[idx][1] = seg[2 * idx + 1][1];
                seg[idx][2] = seg[2 * idx + 2][2];
                if (len1 == (mid - l + 1))
                {
                    seg[idx][1] = len1 + seg[2 * idx + 2][1];
                }
                if (len2 == (r - (mid)))
                {
                    seg[idx][2] = len2 + seg[2 * idx + 1][2];
                }
                return;
            }

            else
            {
                seg[idx][0] = max(mxleft, mxright);
            }
        }
        else
        {
            seg[idx][0] = max(mxleft, mxright);
        }

        seg[idx][1] = seg[2 * idx + 1][1];
        seg[idx][2] = seg[2 * idx + 2][2];
        return;
    }
    void print()
    {
        for (auto i : seg)
            cout << i[0] << " " << i[1] << " " << i[2] << endl;
    }
    void update(int l, int r, int idx, int updateidx, char val)
    {
        if (l == r)
        {
            s[updateidx] = val;
            seg[idx] = {1, 1, 1};
            return;
        }
        int mid = (l + r) / 2;
        if (updateidx <= mid)
        {
            update(l, mid, 2 * idx + 1, updateidx, val);
        }
        else
        {
            update(mid + 1, r, 2 * idx + 2, updateidx, val);
        }
        int len1 = seg[2 * idx + 1][2];
        int len2 = seg[2 * idx + 2][1];
        int mxleft = seg[2 * idx + 1][0];
        int mxright = seg[2 * idx + 2][0];
        // if(idx==1){
        //     cout<<len1<<" "<<len2<<" "<<mxleft<<" "<<mxright<<endl;
        // }
        if ((s[mid] == s[mid + 1]))
        {

            if ((len1 + len2 >= max(mxleft, mxright)))
            {
                seg[idx][0] = len1 + len2;
                seg[idx][1] = seg[2 * idx + 1][1];
                seg[idx][2] = seg[2 * idx + 2][2];
                if (len1 == (mid - l + 1))
                {
                    seg[idx][1] = len1 + seg[2 * idx + 2][1];
                }
                if (len2 == (r - (mid)))
                {
                    seg[idx][2] = len2 + seg[2 * idx + 1][2];
                }
                return;
            }

            else
            {
                seg[idx][0] = max(mxleft, mxright);
            }
        }
        else
        {
            seg[idx][0] = max(mxleft, mxright);
        }

        seg[idx][1] = seg[2 * idx + 1][1];
        seg[idx][2] = seg[2 * idx + 2][2];
        return;
    }
    int query()
    {
        return seg[0][0];
    }
};

class Solution {
public:
  vector<int> longestRepeating(string s, string queryCharacters, vector<int> &queryIndices)
{
    vector<int> ans;
    SegmentTree segmentTree(s.length(), s);
    int n = s.length();
    segmentTree.build(0, n - 1, 0);

    for (int i = 0; i < queryCharacters.size(); i++)
    {
        segmentTree.update(0, n - 1, 0, queryIndices[i], queryCharacters[i]);
        ans.push_back(segmentTree.query());
    }

    // for (auto i : ans)
    //     cout << i << " ";
    return ans;
}
};
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