Longest Substring of One Repeating Character
<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>
Segment Tree; Array; String; Ordered Set
type Node struct {
	l, r         int
	lmx, rmx, mx int
}

type SegmentTree struct {
	s  []byte
	tr []*Node
}

func NewNode(l, r int) *Node {
	return &Node{l: l, r: r, lmx: 1, rmx: 1, mx: 1}
}

func NewSegmentTree(s string) *SegmentTree {
	n := len(s)
	tree := &SegmentTree{s: []byte(s), tr: make([]*Node, n<<2)}
	tree.build(1, 1, n)
	return tree
}

func (tree *SegmentTree) build(u, l, r int) {
	tree.tr[u] = NewNode(l, r)
	if l == r {
		return
	}
	mid := (l + r) >> 1
	tree.build(u<<1, l, mid)
	tree.build(u<<1|1, mid+1, r)
	tree.pushup(u)
}

func (tree *SegmentTree) modify(u, x int, v byte) {
	if tree.tr[u].l == x && tree.tr[u].r == x {
		tree.s[x-1] = v
		return
	}
	mid := (tree.tr[u].l + tree.tr[u].r) >> 1
	if x <= mid {
		tree.modify(u<<1, x, v)
	} else {
		tree.modify(u<<1|1, x, v)
	}
	tree.pushup(u)
}

func (tree *SegmentTree) query(u, l, r int) int {
	if tree.tr[u].l >= l && tree.tr[u].r <= r {
		return tree.tr[u].mx
	}
	mid := (tree.tr[u].l + tree.tr[u].r) >> 1
	ans := 0
	if r <= mid {
		ans = tree.query(u<<1, l, r)
	} else if l > mid {
		ans = max(ans, tree.query(u<<1|1, l, r))
	} else {
		ans = max(tree.query(u<<1, l, r), tree.query(u<<1|1, l, r))
	}
	return ans
}

func (tree *SegmentTree) pushup(u int) {
	root := tree.tr[u]
	left := tree.tr[u<<1]
	right := tree.tr[u<<1|1]
	root.mx = max(left.mx, right.mx)
	root.lmx = left.lmx
	root.rmx = right.rmx
	a := left.r - left.l + 1
	b := right.r - right.l + 1
	if tree.s[left.r-1] == tree.s[right.l-1] {
		if left.lmx == a {
			root.lmx += right.lmx
		}
		if right.rmx == b {
			root.rmx += left.rmx
		}
		root.mx = max(root.mx, left.rmx+right.lmx)
	}
}

func longestRepeating(s string, queryCharacters string, queryIndices []int) (ans []int) {
	tree := NewSegmentTree(s)
	n := len(s)
	for i, v := range queryCharacters {
		x := queryIndices[i] + 1
		tree.modify(1, x, byte(v))
		ans = append(ans, tree.query(1, 1, n))
	}
	return
}