<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:
    def longestRepeating(self, s: str, queryCharacters: str, queryIndices: List[int]) -> List[int]:
        


        #node stores (longest first, longest in general, longest last)
        #also store which letter they are
        #base case is (1, 1, 1)
        #transitions are:
        #new longest = max(2 longest in general, longest last (left) + longest first (right) if they are the same)

        def create(seg_tree, node, l, r):
            if l > r:
                return ((0, "?"), (0, "?"), (0, "?"))
            if l == r:
                seg_tree[node - 1] = ((1, s[l]), (1, s[l]), (1, s[l]))
                return seg_tree[node - 1]

            mid = (l + r) // 2

            lf, lg, ll = create(seg_tree, node * 2, l, mid)
            rf, rg, rl = create(seg_tree, node * 2 + 1, mid + 1, r)

            mf, mg, ml = lf, max(lg, rg), rl

            if ll[1] == rf[1]:
                mg = max((ll[0] + rf[0], rf[1]), mg)

                if lf[0] == mid - l + 1:
                    mf = (lf[0] + rf[0], lf[1])

                if rl[0] == r - mid:
                    ml = (ll[0] + rl[0], rl[1])


            seg_tree[node - 1] = (mf, mg, ml)

            return seg_tree[node - 1]


        def update(seg_tree, node, l, r, ind, val):
            if l > r:
                return ((0, "?"), (0, "?"), (0, "?"))
            if l > ind or r < ind:
                return seg_tree[node - 1]
            if l == r:
                seg_tree[node - 1] = ((1, val), (1, val), (1, val))
                return seg_tree[node - 1]


            mid = (l + r) // 2

            lf, lg, ll = update(seg_tree, node * 2, l, mid, ind, val)
            rf, rg, rl = update(seg_tree, node * 2 + 1, mid + 1, r, ind, val)

            mf, mg, ml = lf, max(lg, rg), rl

            if ll[1] == rf[1]:
                mg = max((ll[0] + rf[0], rf[1]), mg)

                if lf[0] == mid - l + 1:
                    mf = (lf[0] + rf[0], lf[1])

                if rl[0] == r - mid:
                    ml = (ll[0] + rl[0], rl[1])


            seg_tree[node - 1] = (mf, mg, ml)

            return seg_tree[node - 1]



        seg_tree = [0] * len(s) * 4
        ans = []

        create(seg_tree, 1, 0, len(s) - 1)

        for i in range(len(queryCharacters)):
            update(seg_tree, 1, 0, len(s) - 1, queryIndices[i], queryCharacters[i])
            ans.append(seg_tree[0][1][0])

        return ans
4707