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Formatted question description: https://leetcode.ca/all/2213.html

# 2213. Longest Substring of One Repeating Character

• Difficulty: Hard.
• Related Topics: Array, String, Segment Tree, Ordered Set.
• Similar Questions: Merge Intervals, Longest Repeating Character Replacement, Consecutive Characters, Create Sorted Array through Instructions.

## Problem

You are given a 0-indexed string s. You are also given a 0-indexed string queryCharacters of length k and a 0-indexed array of integer indices queryIndices of length k, both of which are used to describe k queries.

The ith query updates the character in s at index queryIndices[i] to the character queryCharacters[i].

Return an array lengths of length **k where** lengths[i] is the **length of the longest substring of s consisting of only one repeating character after the** ith query** is performed.**

Example 1:

Input: s = "babacc", queryCharacters = "bcb", queryIndices = [1,3,3]
Output: [3,3,4]
Explanation:
- 1st query updates s = "bbbacc". The longest substring consisting of one repeating character is "bbb" with length 3.
- 2nd query updates s = "bbbccc".
The longest substring consisting of one repeating character can be "bbb" or "ccc" with length 3.
- 3rd query updates s = "bbbbcc". The longest substring consisting of one repeating character is "bbbb" with length 4.
Thus, we return [3,3,4].


Example 2:

Input: s = "abyzz", queryCharacters = "aa", queryIndices = [2,1]
Output: [2,3]
Explanation:
- 1st query updates s = "abazz". The longest substring consisting of one repeating character is "zz" with length 2.
- 2nd query updates s = "aaazz". The longest substring consisting of one repeating character is "aaa" with length 3.
Thus, we return [2,3].


Constraints:

• 1 <= s.length <= 105

• s consists of lowercase English letters.

• k == queryCharacters.length == queryIndices.length

• 1 <= k <= 105

• queryCharacters consists of lowercase English letters.

• 0 <= queryIndices[i] < s.length

## Solution

• class Solution {
static class TreeNode {
int start;
int end;
char leftChar;
int leftCharLen;
char rightChar;
int rightCharLen;
int max;
TreeNode left;
TreeNode right;

TreeNode(int start, int end) {
this.start = start;
this.end = end;
left = null;
right = null;
}
}

public int[] longestRepeating(String s, String queryCharacters, int[] queryIndices) {
char[] sChar = s.toCharArray();
char[] qChar = queryCharacters.toCharArray();
TreeNode root = buildTree(sChar, 0, sChar.length - 1);
int[] result = new int[qChar.length];
for (int i = 0; i < qChar.length; i++) {
updateTree(root, queryIndices[i], qChar[i]);
if (root != null) {
result[i] = root.max;
}
}
return result;
}

private TreeNode buildTree(char[] s, int from, int to) {
if (from > to) {
return null;
}
TreeNode root = new TreeNode(from, to);
if (from == to) {
root.max = 1;
root.rightChar = root.leftChar = s[from];
root.leftCharLen = root.rightCharLen = 1;
return root;
}
int middle = from + (to - from) / 2;
root.left = buildTree(s, from, middle);
root.right = buildTree(s, middle + 1, to);
updateNode(root);
return root;
}

private void updateTree(TreeNode root, int index, char c) {
if (root == null || root.start > index || root.end < index) {
return;
}
if (root.start == index && root.end == index) {
root.leftChar = root.rightChar = c;
return;
}
updateTree(root.left, index, c);
updateTree(root.right, index, c);
updateNode(root);
}

private void updateNode(TreeNode root) {
if (root == null) {
return;
}
root.leftChar = root.left.leftChar;
root.leftCharLen = root.left.leftCharLen;
root.rightChar = root.right.rightChar;
root.rightCharLen = root.right.rightCharLen;
root.max = Math.max(root.left.max, root.right.max);
if (root.left.rightChar == root.right.leftChar) {
int len = root.left.rightCharLen + root.right.leftCharLen;
if (root.left.leftChar == root.left.rightChar
&& root.left.leftCharLen == root.left.end - root.left.start + 1) {
root.leftCharLen = len;
}
if (root.right.leftChar == root.right.rightChar
&& root.right.leftCharLen == root.right.end - root.right.start + 1) {
root.rightCharLen = len;
}
root.max = Math.max(root.max, len);
}
}
}

############

class Node {
int l;
int r;
int size;
int lmx;
int rmx;
int mx;
char lc;
char rc;
}

class SegmentTree {
private String s;
private Node[] tr;

public SegmentTree(String s) {
int n = s.length();
this.s = s;
tr = new Node[n << 2];
for (int i = 0; i < tr.length; ++i) {
tr[i] = new Node();
}
build(1, 1, n);
}

public void build(int u, int l, int r) {
tr[u].l = l;
tr[u].r = r;
if (l == r) {
tr[u].lmx = 1;
tr[u].rmx = 1;
tr[u].mx = 1;
tr[u].size = 1;
tr[u].lc = s.charAt(l - 1);
tr[u].rc = s.charAt(l - 1);
return;
}
int mid = (l + r) >> 1;
build(u << 1, l, mid);
build(u << 1 | 1, mid + 1, r);
pushup(u);
}

void modify(int u, int x, char v) {
if (tr[u].l == x && tr[u].r == x) {
tr[u].lc = v;
tr[u].rc = v;
return;
}
int mid = (tr[u].l + tr[u].r) >> 1;
if (x <= mid) {
modify(u << 1, x, v);
} else {
modify(u << 1 | 1, x, v);
}
pushup(u);
}

Node query(int u, int l, int r) {
if (tr[u].l >= l && tr[u].r <= r) {
return tr[u];
}
int mid = (tr[u].l + tr[u].r) >> 1;
if (r <= mid) {
return query(u << 1, l, r);
}
if (l > mid) {
return query(u << 1 | 1, l, r);
}
Node ans = new Node();
Node left = query(u << 1, l, r);
Node right = query(u << 1 | 1, l, r);
pushup(ans, left, right);
return ans;
}

void pushup(Node root, Node left, Node right) {
root.lc = left.lc;
root.rc = right.rc;
root.size = left.size + right.size;

root.mx = Math.max(left.mx, right.mx);
root.lmx = left.lmx;
root.rmx = right.rmx;

if (left.rc == right.lc) {
if (left.lmx == left.size) {
root.lmx += right.lmx;
}
if (right.rmx == right.size) {
root.rmx += left.rmx;
}
root.mx = Math.max(root.mx, left.rmx + right.lmx);
}
}

void pushup(int u) {
pushup(tr[u], tr[u << 1], tr[u << 1 | 1]);
}
}

class Solution {
public int[] longestRepeating(String s, String queryCharacters, int[] queryIndices) {
SegmentTree tree = new SegmentTree(s);
int k = queryCharacters.length();
int[] ans = new int[k];
for (int i = 0; i < k; ++i) {
int x = queryIndices[i] + 1;
char c = queryCharacters.charAt(i);
tree.modify(1, x, c);
ans[i] = tree.query(1, 1, s.length()).mx;
}
return ans;
}
}

• class Node:
def __init__(self):
self.l = 0
self.r = 0
self.lmx = 0
self.rmx = 0
self.mx = 0
self.size = 0
self.lc = None
self.rc = None

N = 100010
tr = [Node() for _ in range(N << 2)]

class SegmentTree:
def __init__(self, s):
n = len(s)
self.s = s
self.build(1, 1, n)

def build(self, u, l, r):
tr[u].l = l
tr[u].r = r
if l == r:
tr[u].lmx = tr[u].rmx = tr[u].mx = tr[u].size = 1
tr[u].lc = tr[u].rc = self.s[l - 1]
return
mid = (l + r) >> 1
self.build(u << 1, l, mid)
self.build(u << 1 | 1, mid + 1, r)
self.pushup(u)

def modify(self, u, x, v):
if tr[u].l == x and tr[u].r == x:
tr[u].lc = tr[u].rc = v
return
mid = (tr[u].l + tr[u].r) >> 1
if x <= mid:
self.modify(u << 1, x, v)
else:
self.modify(u << 1 | 1, x, v)
self.pushup(u)

def query(self, u, l, r):
if tr[u].l >= l and tr[u].r <= r:
return tr[u]
mid = (tr[u].l + tr[u].r) >> 1
if r <= mid:
return self.query(u << 1, l, r)
if l > mid:
return self.query(u << 1 | 1, l, r)
left, right = self.query(u << 1, l, r), self.query(u << 1 | 1, l, r)
ans = Node()
self._pushup(ans, left, right)
return ans

def _pushup(self, root, left, right):
root.lc, root.rc = left.lc, right.rc
root.size = left.size + right.size

root.mx = max(left.mx, right.mx)
root.lmx, root.rmx = left.lmx, right.rmx

if left.rc == right.lc:
if left.lmx == left.size:
root.lmx += right.lmx
if right.rmx == right.size:
root.rmx += left.rmx
root.mx = max(root.mx, left.rmx + right.lmx)

def pushup(self, u):
self._pushup(tr[u], tr[u << 1], tr[u << 1 | 1])

class Solution:
def longestRepeating(
self, s: str, queryCharacters: str, queryIndices: List[int]
) -> List[int]:
tree = SegmentTree(s)
k = len(queryIndices)
ans = []
for i, c in enumerate(queryCharacters):
x = queryIndices[i] + 1
tree.modify(1, x, c)
ans.append(tree.query(1, 1, len(s)).mx)
return ans

############

# 2213. Longest Substring of One Repeating Character
# https://leetcode.com/problems/longest-substring-of-one-repeating-character/

from sortedcontainers import SortedList

class Solution:
def longestRepeating(self, s: str, qc: str, qi: List[int]) -> List[int]:
M = 10 ** 6
n = len(s)
k = len(qc)

s = list(s)

sl = SortedList()
h = SortedList()

curr = 0

for x, group in groupby(s):
length = sum(1 for c in group)
sl.add((curr, curr + length - 1))
h.add(length)
curr += length

def add(x, y):
sl.add((x, y))
h.add(y - x + 1)

def remove(x, y):
sl.remove((x, y))
h.remove(y - x + 1)

def split(i):
index = sl.bisect_right((i, M)) - 1

left, right = sl[index]
remove(left, right)

if left != i:
add(left, i - 1)

add(i, i)

if right != i:
add(i + 1, right)

def merge(index):
if index + 1 >= len(sl): return

left1, right1 = sl[index]
left2, right2 = sl[index + 1]

if s[right1] != s[left2]: return

remove(left1, right1)
remove(left2, right2)
add(left1, right2)

res = []

for i, c in zip(qi, qc):
if s[i] == c:
res.append(h[-1])
continue

split(i)

index = sl.bisect_right((i, M)) - 1

s[i] = c
merge(index)

if index - 1 >= 0:
merge(index - 1)

res.append(h[-1])

return res


• class Node {
public:
int l, r, size, lmx, rmx, mx;
char lc, rc;
};

class SegmentTree {
private:
string s;
vector<Node*> tr;

public:
SegmentTree(string& s) {
this->s = s;
int n = s.size();
tr.resize(n << 2);
for (int i = 0; i < tr.size(); ++i) tr[i] = new Node();
build(1, 1, n);
}

void build(int u, int l, int r) {
tr[u]->l = l;
tr[u]->r = r;
if (l == r) {
tr[u]->lmx = tr[u]->rmx = tr[u]->mx = tr[u]->size = 1;
tr[u]->lc = tr[u]->rc = s[l - 1];
return;
}
int mid = (l + r) >> 1;
build(u << 1, l, mid);
build(u << 1 | 1, mid + 1, r);
pushup(u);
}

void modify(int u, int x, char v) {
if (tr[u]->l == x && tr[u]->r == x) {
tr[u]->lc = tr[u]->rc = v;
return;
}
int mid = (tr[u]->l + tr[u]->r) >> 1;
if (x <= mid)
modify(u << 1, x, v);
else
modify(u << 1 | 1, x, v);
pushup(u);
}

Node* query(int u, int l, int r) {
if (tr[u]->l >= l && tr[u]->r <= r) return tr[u];
int mid = (tr[u]->l + tr[u]->r) >> 1;
if (r <= mid) return query(u << 1, l, r);
if (l > mid) query(u << 1 | 1, l, r);
Node* ans = new Node();
Node* left = query(u << 1, l, r);
Node* right = query(u << 1 | 1, l, r);
pushup(ans, left, right);
return ans;
}

void pushup(Node* root, Node* left, Node* right) {
root->lc = left->lc;
root->rc = right->rc;
root->size = left->size + right->size;

root->mx = max(left->mx, right->mx);
root->lmx = left->lmx;
root->rmx = right->rmx;

if (left->rc == right->lc) {
if (left->lmx == left->size) root->lmx += right->lmx;
if (right->rmx == right->size) root->rmx += left->rmx;
root->mx = max(root->mx, left->rmx + right->lmx);
}
}

void pushup(int u) {
pushup(tr[u], tr[u << 1], tr[u << 1 | 1]);
}
};

class Solution {
public:
vector<int> longestRepeating(string s, string queryCharacters, vector<int>& queryIndices) {
SegmentTree* tree = new SegmentTree(s);
int k = queryCharacters.size();
vector<int> ans(k);
for (int i = 0; i < k; ++i) {
int x = queryIndices[i] + 1;
tree->modify(1, x, queryCharacters[i]);
ans[i] = tree->query(1, 1, s.size())->mx;
}
return ans;
}
};

• type segmentTree struct {
str []byte
mx  []int
lmx []int
rmx []int
}

func newSegmentTree(s string) *segmentTree {
n := len(s)
t := &segmentTree{
str: []byte(s),
mx:  make([]int, n<<2),
lmx: make([]int, n<<2),
rmx: make([]int, n<<2),
}
t.build(0, 0, n-1)
return t
}

func (t *segmentTree) build(x, l, r int) {
if l == r {
t.lmx[x] = 1
t.rmx[x] = 1
t.mx[x] = 1
return
}
m := int(uint(l+r) >> 1)
t.build(x*2+1, l, m)
t.build(x*2+2, m+1, r)
t.pushup(x, l, m, r)
}

func (t *segmentTree) pushup(x, l, m, r int) {
lch, rch := x*2+1, x*2+2
t.lmx[x] = t.lmx[lch]
t.rmx[x] = t.rmx[rch]
t.mx[x] = max(t.mx[lch], t.mx[rch])
// can be merged
if t.str[m] == t.str[m+1] {
if t.lmx[lch] == m-l+1 {
t.lmx[x] += t.lmx[rch]
}
if t.rmx[rch] == r-m {
t.rmx[x] += t.rmx[lch]
}
t.mx[x] = max(t.mx[x], t.rmx[lch]+t.lmx[rch])
}
}

func (t *segmentTree) update(x, l, r, pos int, val byte) {
if l == r {
t.str[pos] = val
return
}
m := int(uint(l+r) >> 1)
if pos <= m {
t.update(x*2+1, l, m, pos, val)
} else {
t.update(x*2+2, m+1, r, pos, val)
}
t.pushup(x, l, m, r)
}

func max(x, y int) int {
if x > y {
return x
}
return y
}

func longestRepeating(s string, queryCharacters string, queryIndices []int) []int {
ans := make([]int, len(queryCharacters))
t := newSegmentTree(s)
n := len(s)
for i, c := range queryCharacters {
t.update(0, 0, n-1, queryIndices[i], byte(c))
ans[i] = t.mx[0]
}
return ans
}



Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).