2612. Minimum Reverse Operations
Description
You are given an integer n and an integer p representing an array arr of length n where all elements are set to 0's, except position p which is set to 1. You are also given an integer array banned containing restricted positions. Perform the following operation on arr:
- Reverse a subarray with size
kif the single 1 is not set to a position inbanned.
Return an integer array answer with n results where the ith result is the minimum number of operations needed to bring the single 1 to position i in arr, or -1 if it is impossible.
Example 1:
Input: n = 4, p = 0, banned = [1,2], k = 4
Output: [0,-1,-1,1]
Explanation:
- Initially 1 is placed at position 0 so the number of operations we need for position 0 is 0.
- We can never place 1 on the banned positions, so the answer for positions 1 and 2 is -1.
- Perform the operation of size 4 to reverse the whole array.
- After a single operation 1 is at position 3 so the answer for position 3 is 1.
Example 2:
Input: n = 5, p = 0, banned = [2,4], k = 3
Output: [0,-1,-1,-1,-1]
Explanation:
- Initially 1 is placed at position 0 so the number of operations we need for position 0 is 0.
- We cannot perform the operation on the subarray positions
[0, 2]because position 2 is in banned. - Because 1 cannot be set at position 2, it is impossible to set 1 at other positions in more operations.
Example 3:
Input: n = 4, p = 2, banned = [0,1,3], k = 1
Output: [-1,-1,0,-1]
Explanation:
Perform operations of size 1 and 1 never changes its position.
Constraints:
1 <= n <= 1050 <= p <= n - 10 <= banned.length <= n - 10 <= banned[i] <= n - 11 <= k <= nbanned[i] != p- all values in
bannedare unique
Solutions
Solution 1: Ordered Set + BFS
We notice that for any index $i$ in the subarray interval $[l,..r]$, the flipped index $j = l + r - i$.
If the subarray moves one position to the right, then $j = l + 1 + r + 1 - i = l + r - i + 2$, that is, $j$ will increase by $2$.
Similarly, if the subarray moves one position to the left, then $j = l - 1 + r - 1 - i = l + r - i - 2$, that is, $j$ will decrease by $2$.
Therefore, for a specific index $i$, all its flipped indices form an arithmetic progression with common difference $2$, that is, all the flipped indices have the same parity.
Next, we consider the range of values of the index $i$ after flipping $j$.
If the boundary is not considered, the range of values of $j$ is $[i - k + 1, i + k - 1]$.
If the subarray is on the left, then $[l, r] = [0, k - 1]$, so the flipped index $j$ of $i$ is $0 + k - 1 - i$, that is, $j = k - i - 1$, so the left boundary $mi = max(i - k + 1, k - i - 1)$.
If the subarray is on the right, then $[l, r] = [n - k, n - 1]$, so the flipped index $j= n - k + n - 1 - i$ is $j = n \times 2 - k - i - 1$, so the right boundary of $j$ is $mx = min(i + k - 1, n \times 2 - k - i - 1)$.
We use two ordered sets to store all the odd indices and even indices to be searched, here we need to exclude the indices in the array $banned$ and the index $p$.
Then we use BFS to search, each time searching all the flipped indices $j$ of the current index $i$, that is, $j = mi, mi + 2, mi + 4, \dots, mx$, updating the answer of index $j$ and adding index $j$ to the search queue, and removing index $j$ from the corresponding ordered set.
When the search is over, the answer to all indices can be obtained.
The time complexity is $O(n \times \log n)$ and the space complexity is $O(n)$. Where $n$ is the given array length in the problem.
Python3
class Solution:
def minReverseOperations(
self, n: int, p: int, banned: List[int], k: int
) -> List[int]:
ans = [-1] * n
ans[p] = 0
ts = [SortedSet() for _ in range(2)]
for i in range(n):
ts[i % 2].add(i)
ts[p % 2].remove(p)
for i in banned:
ts[i % 2].remove(i)
ts[0].add(n)
ts[1].add(n)
q = deque([p])
while q:
i = q.popleft()
mi = max(i - k + 1, k - i - 1)
mx = min(i + k - 1, n * 2 - k - i - 1)
s = ts[mi % 2]
j = s.bisect_left(mi)
while s[j] <= mx:
q.append(s[j])
ans[s[j]] = ans[i] + 1
s.remove(s[j])
j = s.bisect_left(mi)
return ans
Java
class Solution {
public int[] minReverseOperations(int n, int p, int[] banned, int k) {
int[] ans = new int[n];
TreeSet<Integer>[] ts = new TreeSet[] {new TreeSet<>(), new TreeSet<>()};
for (int i = 0; i < n; ++i) {
ts[i % 2].add(i);
ans[i] = i == p ? 0 : -1;
}
ts[p % 2].remove(p);
for (int i : banned) {
ts[i % 2].remove(i);
}
ts[0].add(n);
ts[1].add(n);
Deque<Integer> q = new ArrayDeque<>();
q.offer(p);
while (!q.isEmpty()) {
int i = q.poll();
int mi = Math.max(i - k + 1, k - i - 1);
int mx = Math.min(i + k - 1, n * 2 - k - i - 1);
var s = ts[mi % 2];
for (int j = s.ceiling(mi); j <= mx; j = s.ceiling(mi)) {
q.offer(j);
ans[j] = ans[i] + 1;
s.remove(j);
}
}
return ans;
}
}
C++
class Solution {
public:
vector<int> minReverseOperations(int n, int p, vector<int>& banned, int k) {
vector<int> ans(n, -1);
ans[p] = 0;
set<int> ts[2];
for (int i = 0; i < n; ++i) {
ts[i % 2].insert(i);
}
ts[p % 2].erase(p);
for (int i : banned) {
ts[i % 2].erase(i);
}
ts[0].insert(n);
ts[1].insert(n);
queue<int> q{{p}};
while (!q.empty()) {
int i = q.front();
q.pop();
int mi = max(i - k + 1, k - i - 1);
int mx = min(i + k - 1, n * 2 - k - i - 1);
auto& s = ts[mi % 2];
auto it = s.lower_bound(mi);
while (*it <= mx) {
int j = *it;
ans[j] = ans[i] + 1;
q.push(j);
it = s.erase(it);
}
}
return ans;
}
};
Go
func minReverseOperations(n int, p int, banned []int, k int) []int {
ans := make([]int, n)
ts := [2]*redblacktree.Tree{redblacktree.NewWithIntComparator(), redblacktree.NewWithIntComparator()}
for i := 0; i < n; i++ {
ts[i%2].Put(i, struct{}{})
ans[i] = -1
}
ans[p] = 0
ts[p%2].Remove(p)
for _, i := range banned {
ts[i%2].Remove(i)
}
ts[0].Put(n, struct{}{})
ts[1].Put(n, struct{}{})
q := []int{p}
for len(q) > 0 {
i := q[0]
q = q[1:]
mi := max(i-k+1, k-i-1)
mx := min(i+k-1, n*2-k-i-1)
s := ts[mi%2]
for x, _ := s.Ceiling(mi); x.Key.(int) <= mx; x, _ = s.Ceiling(mi) {
j := x.Key.(int)
s.Remove(j)
ans[j] = ans[i] + 1
q = append(q, j)
}
}
return ans
}
TypeScript
function minReverseOperations(n: number, p: number, banned: number[], k: number): number[] {
const ans: number[] = Array(n).fill(-1);
const ts = [new TreeSet<number>(), new TreeSet<number>()];
for (let i = 0; i < n; ++i) {
ts[i % 2].add(i);
}
ans[p] = 0;
ts[p % 2].delete(p);
for (const i of banned) {
ts[i % 2].delete(i);
}
ts[0].add(n);
ts[1].add(n);
let q = [p];
while (q.length) {
const t: number[] = [];
for (const i of q) {
const mi = Math.max(i - k + 1, k - i - 1);
const mx = Math.min(i + k - 1, n * 2 - k - i - 1);
const s = ts[mi % 2];
for (let j = s.ceil(mi)!; j <= mx; j = s.ceil(j)!) {
t.push(j);
ans[j] = ans[i] + 1;
s.delete(j);
}
}
q = t;
}
return ans;
}
type Compare<T> = (lhs: T, rhs: T) => number;
class RBTreeNode<T = number> {
data: T;
count: number;
left: RBTreeNode<T> | null;
right: RBTreeNode<T> | null;
parent: RBTreeNode<T> | null;
color: number;
constructor(data: T) {
this.data = data;
this.left = this.right = this.parent = null;
this.color = 0;
this.count = 1;
}
sibling(): RBTreeNode<T> | null {
if (!this.parent) return null; // sibling null if no parent
return this.isOnLeft() ? this.parent.right : this.parent.left;
}
isOnLeft(): boolean {
return this === this.parent!.left;
}
hasRedChild(): boolean {
return (
Boolean(this.left && this.left.color === 0) ||
Boolean(this.right && this.right.color === 0)
);
}
}
class RBTree<T> {
root: RBTreeNode<T> | null;
lt: (l: T, r: T) => boolean;
constructor(compare: Compare<T> = (l: T, r: T) => (l < r ? -1 : l > r ? 1 : 0)) {
this.root = null;
this.lt = (l: T, r: T) => compare(l, r) < 0;
}
rotateLeft(pt: RBTreeNode<T>): void {
const right = pt.right!;
pt.right = right.left;
if (pt.right) pt.right.parent = pt;
right.parent = pt.parent;
if (!pt.parent) this.root = right;
else if (pt === pt.parent.left) pt.parent.left = right;
else pt.parent.right = right;
right.left = pt;
pt.parent = right;
}
rotateRight(pt: RBTreeNode<T>): void {
const left = pt.left!;
pt.left = left.right;
if (pt.left) pt.left.parent = pt;
left.parent = pt.parent;
if (!pt.parent) this.root = left;
else if (pt === pt.parent.left) pt.parent.left = left;
else pt.parent.right = left;
left.right = pt;
pt.parent = left;
}
swapColor(p1: RBTreeNode<T>, p2: RBTreeNode<T>): void {
const tmp = p1.color;
p1.color = p2.color;
p2.color = tmp;
}
swapData(p1: RBTreeNode<T>, p2: RBTreeNode<T>): void {
const tmp = p1.data;
p1.data = p2.data;
p2.data = tmp;
}
fixAfterInsert(pt: RBTreeNode<T>): void {
let parent = null;
let grandParent = null;
while (pt !== this.root && pt.color !== 1 && pt.parent?.color === 0) {
parent = pt.parent;
grandParent = pt.parent.parent;
/* Case : A
Parent of pt is left child of Grand-parent of pt */
if (parent === grandParent?.left) {
const uncle = grandParent.right;
/* Case : 1
The uncle of pt is also red
Only Recoloring required */
if (uncle && uncle.color === 0) {
grandParent.color = 0;
parent.color = 1;
uncle.color = 1;
pt = grandParent;
} else {
/* Case : 2
pt is right child of its parent
Left-rotation required */
if (pt === parent.right) {
this.rotateLeft(parent);
pt = parent;
parent = pt.parent;
}
/* Case : 3
pt is left child of its parent
Right-rotation required */
this.rotateRight(grandParent);
this.swapColor(parent!, grandParent);
pt = parent!;
}
} else {
/* Case : B
Parent of pt is right child of Grand-parent of pt */
const uncle = grandParent!.left;
/* Case : 1
The uncle of pt is also red
Only Recoloring required */
if (uncle != null && uncle.color === 0) {
grandParent!.color = 0;
parent.color = 1;
uncle.color = 1;
pt = grandParent!;
} else {
/* Case : 2
pt is left child of its parent
Right-rotation required */
if (pt === parent.left) {
this.rotateRight(parent);
pt = parent;
parent = pt.parent;
}
/* Case : 3
pt is right child of its parent
Left-rotation required */
this.rotateLeft(grandParent!);
this.swapColor(parent!, grandParent!);
pt = parent!;
}
}
}
this.root!.color = 1;
}
delete(val: T): boolean {
const node = this.find(val);
if (!node) return false;
node.count--;
if (!node.count) this.deleteNode(node);
return true;
}
deleteAll(val: T): boolean {
const node = this.find(val);
if (!node) return false;
this.deleteNode(node);
return true;
}
deleteNode(v: RBTreeNode<T>): void {
const u = BSTreplace(v);
// True when u and v are both black
const uvBlack = (u === null || u.color === 1) && v.color === 1;
const parent = v.parent!;
if (!u) {
// u is null therefore v is leaf
if (v === this.root) this.root = null;
// v is root, making root null
else {
if (uvBlack) {
// u and v both black
// v is leaf, fix double black at v
this.fixDoubleBlack(v);
} else {
// u or v is red
if (v.sibling()) {
// sibling is not null, make it red"
v.sibling()!.color = 0;
}
}
// delete v from the tree
if (v.isOnLeft()) parent.left = null;
else parent.right = null;
}
return;
}
if (!v.left || !v.right) {
// v has 1 child
if (v === this.root) {
// v is root, assign the value of u to v, and delete u
v.data = u.data;
v.left = v.right = null;
} else {
// Detach v from tree and move u up
if (v.isOnLeft()) parent.left = u;
else parent.right = u;
u.parent = parent;
if (uvBlack) this.fixDoubleBlack(u);
// u and v both black, fix double black at u
else u.color = 1; // u or v red, color u black
}
return;
}
// v has 2 children, swap data with successor and recurse
this.swapData(u, v);
this.deleteNode(u);
// find node that replaces a deleted node in BST
function BSTreplace(x: RBTreeNode<T>): RBTreeNode<T> | null {
// when node have 2 children
if (x.left && x.right) return successor(x.right);
// when leaf
if (!x.left && !x.right) return null;
// when single child
return x.left ?? x.right;
}
// find node that do not have a left child
// in the subtree of the given node
function successor(x: RBTreeNode<T>): RBTreeNode<T> {
let temp = x;
while (temp.left) temp = temp.left;
return temp;
}
}
fixDoubleBlack(x: RBTreeNode<T>): void {
if (x === this.root) return; // Reached root
const sibling = x.sibling();
const parent = x.parent!;
if (!sibling) {
// No sibiling, double black pushed up
this.fixDoubleBlack(parent);
} else {
if (sibling.color === 0) {
// Sibling red
parent.color = 0;
sibling.color = 1;
if (sibling.isOnLeft()) this.rotateRight(parent);
// left case
else this.rotateLeft(parent); // right case
this.fixDoubleBlack(x);
} else {
// Sibling black
if (sibling.hasRedChild()) {
// at least 1 red children
if (sibling.left && sibling.left.color === 0) {
if (sibling.isOnLeft()) {
// left left
sibling.left.color = sibling.color;
sibling.color = parent.color;
this.rotateRight(parent);
} else {
// right left
sibling.left.color = parent.color;
this.rotateRight(sibling);
this.rotateLeft(parent);
}
} else {
if (sibling.isOnLeft()) {
// left right
sibling.right!.color = parent.color;
this.rotateLeft(sibling);
this.rotateRight(parent);
} else {
// right right
sibling.right!.color = sibling.color;
sibling.color = parent.color;
this.rotateLeft(parent);
}
}
parent.color = 1;
} else {
// 2 black children
sibling.color = 0;
if (parent.color === 1) this.fixDoubleBlack(parent);
else parent.color = 1;
}
}
}
}
insert(data: T): boolean {
// search for a position to insert
let parent = this.root;
while (parent) {
if (this.lt(data, parent.data)) {
if (!parent.left) break;
else parent = parent.left;
} else if (this.lt(parent.data, data)) {
if (!parent.right) break;
else parent = parent.right;
} else break;
}
// insert node into parent
const node = new RBTreeNode(data);
if (!parent) this.root = node;
else if (this.lt(node.data, parent.data)) parent.left = node;
else if (this.lt(parent.data, node.data)) parent.right = node;
else {
parent.count++;
return false;
}
node.parent = parent;
this.fixAfterInsert(node);
return true;
}
find(data: T): RBTreeNode<T> | null {
let p = this.root;
while (p) {
if (this.lt(data, p.data)) {
p = p.left;
} else if (this.lt(p.data, data)) {
p = p.right;
} else break;
}
return p ?? null;
}
*inOrder(root: RBTreeNode<T> = this.root!): Generator<T, undefined, void> {
if (!root) return;
for (const v of this.inOrder(root.left!)) yield v;
yield root.data;
for (const v of this.inOrder(root.right!)) yield v;
}
*reverseInOrder(root: RBTreeNode<T> = this.root!): Generator<T, undefined, void> {
if (!root) return;
for (const v of this.reverseInOrder(root.right!)) yield v;
yield root.data;
for (const v of this.reverseInOrder(root.left!)) yield v;
}
}
class TreeSet<T = number> {
_size: number;
tree: RBTree<T>;
compare: Compare<T>;
constructor(
collection: T[] | Compare<T> = [],
compare: Compare<T> = (l: T, r: T) => (l < r ? -1 : l > r ? 1 : 0),
) {
if (typeof collection === 'function') {
compare = collection;
collection = [];
}
this._size = 0;
this.compare = compare;
this.tree = new RBTree(compare);
for (const val of collection) this.add(val);
}
size(): number {
return this._size;
}
has(val: T): boolean {
return !!this.tree.find(val);
}
add(val: T): boolean {
const successful = this.tree.insert(val);
this._size += successful ? 1 : 0;
return successful;
}
delete(val: T): boolean {
const deleted = this.tree.deleteAll(val);
this._size -= deleted ? 1 : 0;
return deleted;
}
ceil(val: T): T | undefined {
let p = this.tree.root;
let higher = null;
while (p) {
if (this.compare(p.data, val) >= 0) {
higher = p;
p = p.left;
} else {
p = p.right;
}
}
return higher?.data;
}
floor(val: T): T | undefined {
let p = this.tree.root;
let lower = null;
while (p) {
if (this.compare(val, p.data) >= 0) {
lower = p;
p = p.right;
} else {
p = p.left;
}
}
return lower?.data;
}
higher(val: T): T | undefined {
let p = this.tree.root;
let higher = null;
while (p) {
if (this.compare(val, p.data) < 0) {
higher = p;
p = p.left;
} else {
p = p.right;
}
}
return higher?.data;
}
lower(val: T): T | undefined {
let p = this.tree.root;
let lower = null;
while (p) {
if (this.compare(p.data, val) < 0) {
lower = p;
p = p.right;
} else {
p = p.left;
}
}
return lower?.data;
}
first(): T | undefined {
return this.tree.inOrder().next().value;
}
last(): T | undefined {
return this.tree.reverseInOrder().next().value;
}
shift(): T | undefined {
const first = this.first();
if (first === undefined) return undefined;
this.delete(first);
return first;
}
pop(): T | undefined {
const last = this.last();
if (last === undefined) return undefined;
this.delete(last);
return last;
}
*[Symbol.iterator](): Generator<T, void, void> {
for (const val of this.values()) yield val;
}
*keys(): Generator<T, void, void> {
for (const val of this.values()) yield val;
}
*values(): Generator<T, undefined, void> {
for (const val of this.tree.inOrder()) yield val;
return undefined;
}
/**
* Return a generator for reverse order traversing the set
*/
*rvalues(): Generator<T, undefined, void> {
for (const val of this.tree.reverseInOrder()) yield val;
return undefined;
}
}
class TreeMultiSet<T = number> {
_size: number;
tree: RBTree<T>;
compare: Compare<T>;
constructor(
collection: T[] | Compare<T> = [],
compare: Compare<T> = (l: T, r: T) => (l < r ? -1 : l > r ? 1 : 0),
) {
if (typeof collection === 'function') {
compare = collection;
collection = [];
}
this._size = 0;
this.compare = compare;
this.tree = new RBTree(compare);
for (const val of collection) this.add(val);
}
size(): number {
return this._size;
}
has(val: T): boolean {
return !!this.tree.find(val);
}
add(val: T): boolean {
const successful = this.tree.insert(val);
this._size++;
return successful;
}
delete(val: T): boolean {
const successful = this.tree.delete(val);
if (!successful) return false;
this._size--;
return true;
}
count(val: T): number {
const node = this.tree.find(val);
return node ? node.count : 0;
}
ceil(val: T): T | undefined {
let p = this.tree.root;
let higher = null;
while (p) {
if (this.compare(p.data, val) >= 0) {
higher = p;
p = p.left;
} else {
p = p.right;
}
}
return higher?.data;
}
floor(val: T): T | undefined {
let p = this.tree.root;
let lower = null;
while (p) {
if (this.compare(val, p.data) >= 0) {
lower = p;
p = p.right;
} else {
p = p.left;
}
}
return lower?.data;
}
higher(val: T): T | undefined {
let p = this.tree.root;
let higher = null;
while (p) {
if (this.compare(val, p.data) < 0) {
higher = p;
p = p.left;
} else {
p = p.right;
}
}
return higher?.data;
}
lower(val: T): T | undefined {
let p = this.tree.root;
let lower = null;
while (p) {
if (this.compare(p.data, val) < 0) {
lower = p;
p = p.right;
} else {
p = p.left;
}
}
return lower?.data;
}
first(): T | undefined {
return this.tree.inOrder().next().value;
}
last(): T | undefined {
return this.tree.reverseInOrder().next().value;
}
shift(): T | undefined {
const first = this.first();
if (first === undefined) return undefined;
this.delete(first);
return first;
}
pop(): T | undefined {
const last = this.last();
if (last === undefined) return undefined;
this.delete(last);
return last;
}
*[Symbol.iterator](): Generator<T, void, void> {
yield* this.values();
}
*keys(): Generator<T, void, void> {
for (const val of this.values()) yield val;
}
*values(): Generator<T, undefined, void> {
for (const val of this.tree.inOrder()) {
let count = this.count(val);
while (count--) yield val;
}
return undefined;
}
/**
* Return a generator for reverse order traversing the multi-set
*/
*rvalues(): Generator<T, undefined, void> {
for (const val of this.tree.reverseInOrder()) {
let count = this.count(val);
while (count--) yield val;
}
return undefined;
}
}
Rust
use std::collections::{BTreeSet, VecDeque};
impl Solution {
pub fn min_reverse_operations(n: i32, p: i32, banned: Vec<i32>, k: i32) -> Vec<i32> {
let mut ans = vec![-1; n as usize];
let mut ts = [BTreeSet::new(), BTreeSet::new()];
for i in 0..n {
ts[(i % 2) as usize].insert(i);
}
ans[p as usize] = 0;
ts[(p % 2) as usize].remove(&p);
for &b in &banned {
ts[(b % 2) as usize].remove(&b);
}
ts[0].insert(n);
ts[1].insert(n);
let mut q = VecDeque::new();
q.push_back(p);
while let Some(i) = q.pop_front() {
let mi = (i - k + 1).max(k - i - 1);
let mx = (i + k - 1).min(2 * n - k - i - 1);
let s = &mut ts[(mi % 2) as usize];
while let Some(&j) = s.range(mi..=mx).next() {
q.push_back(j);
ans[j as usize] = ans[i as usize] + 1;
s.remove(&j);
}
}
ans
}
}