1296. Divide Array in Sets of K Consecutive Numbers
Description
Given an array of integers nums and a positive integer k, check whether it is possible to divide this array into sets of k consecutive numbers.
Return true if it is possible. Otherwise, return false.
Example 1:
Input: nums = [1,2,3,3,4,4,5,6], k = 4 Output: true Explanation: Array can be divided into [1,2,3,4] and [3,4,5,6].
Example 2:
Input: nums = [3,2,1,2,3,4,3,4,5,9,10,11], k = 3 Output: true Explanation: Array can be divided into [1,2,3] , [2,3,4] , [3,4,5] and [9,10,11].
Example 3:
Input: nums = [1,2,3,4], k = 3 Output: false Explanation: Each array should be divided in subarrays of size 3.
Constraints:
1 <= k <= nums.length <= 1051 <= nums[i] <= 109
Note: This question is the same as 846: https://leetcode.com/problems/hand-of-straights/
Solutions
Solution 1: Hash Table + Sorting
First, we check if the length of the array $\textit{nums}$ is divisible by $\textit{k}$. If it is not divisible, it means the array cannot be divided into subarrays of length $\textit{k}$, and we return $\text{false}$ directly.
Next, we use a hash table $\textit{cnt}$ to count the occurrences of each number in the array $\textit{nums}$, and then we sort the array $\textit{nums}$.
After sorting, we iterate through the array $\textit{nums}$, and for each number $x$, if $\textit{cnt}[x]$ is not $0$, we enumerate each number $y$ from $x$ to $x + \textit{k} - 1$. If $\textit{cnt}[y]$ is $0$, it means we cannot divide the array into subarrays of length $\textit{k}$, and we return $\text{false}$ directly. Otherwise, we decrement $\textit{cnt}[y]$ by $1$.
After the loop, if no issues were encountered, it means we can divide the array into subarrays of length $\textit{k}$, and we return $\text{true}$.
The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{nums}$.
Python3
class Solution:
def isPossibleDivide(self, nums: List[int], k: int) -> bool:
if len(nums) % k:
return False
cnt = Counter(nums)
for x in sorted(nums):
if cnt[x]:
for y in range(x, x + k):
if cnt[y] == 0:
return False
cnt[y] -= 1
return True
Java
class Solution {
public boolean isPossibleDivide(int[] nums, int k) {
if (nums.length % k != 0) {
return false;
}
Arrays.sort(nums);
Map<Integer, Integer> cnt = new HashMap<>();
for (int x : nums) {
cnt.merge(x, 1, Integer::sum);
}
for (int x : nums) {
if (cnt.getOrDefault(x, 0) > 0) {
for (int y = x; y < x + k; ++y) {
if (cnt.merge(y, -1, Integer::sum) < 0) {
return false;
}
}
}
}
return true;
}
}
C++
class Solution {
public:
bool isPossibleDivide(vector<int>& nums, int k) {
if (nums.size() % k) {
return false;
}
ranges::sort(nums);
unordered_map<int, int> cnt;
for (int x : nums) {
++cnt[x];
}
for (int x : nums) {
if (cnt.contains(x)) {
for (int y = x; y < x + k; ++y) {
if (!cnt.contains(y)) {
return false;
}
if (--cnt[y] == 0) {
cnt.erase(y);
}
}
}
}
return true;
}
};
Go
func isPossibleDivide(nums []int, k int) bool {
if len(nums)%k != 0 {
return false
}
sort.Ints(nums)
cnt := map[int]int{}
for _, x := range nums {
cnt[x]++
}
for _, x := range nums {
if cnt[x] > 0 {
for y := x; y < x+k; y++ {
if cnt[y] == 0 {
return false
}
cnt[y]--
}
}
}
return true
}
TypeScript
function isPossibleDivide(nums: number[], k: number): boolean {
if (nums.length % k !== 0) {
return false;
}
const cnt = new Map<number, number>();
for (const x of nums) {
cnt.set(x, (cnt.get(x) || 0) + 1);
}
nums.sort((a, b) => a - b);
for (const x of nums) {
if (cnt.get(x)! > 0) {
for (let y = x; y < x + k; y++) {
if ((cnt.get(y) || 0) === 0) {
return false;
}
cnt.set(y, cnt.get(y)! - 1);
}
}
}
return true;
}
Solution 2: Ordered Set
Similar to Solution 1, we first check if the length of the array $\textit{nums}$ is divisible by $\textit{k}$. If it is not divisible, it means the array cannot be divided into subarrays of length $\textit{k}$, and we return $\text{false}$ directly.
Next, we use an ordered set $\textit{sd}$ to count the occurrences of each number in the array $\textit{nums}$.
Then, we repeatedly extract the smallest value $x$ from the ordered set and enumerate each number $y$ from $x$ to $x + \textit{k} - 1$. If these numbers all appear in the ordered set with non-zero occurrences, we decrement their occurrence count by $1$. If the occurrence count becomes $0$ after the decrement, we remove the number from the ordered set; otherwise, it means we cannot divide the array into subarrays of length $\textit{k}$, and we return $\text{false}$.
If we can successfully divide the array into subarrays of length $\textit{k}$, we return $\text{true}$ after completing the traversal.
The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{nums}$.
Python3
class Solution:
def isPossibleDivide(self, nums: List[int], k: int) -> bool:
if len(nums) % k:
return False
cnt = Counter(nums)
sd = SortedDict(cnt)
while sd:
x = next(iter(sd))
for y in range(x, x + k):
if y not in sd:
return False
if sd[y] == 1:
del sd[y]
else:
sd[y] -= 1
return True
Java
class Solution {
public boolean isPossibleDivide(int[] nums, int k) {
if (nums.length % k != 0) {
return false;
}
TreeMap<Integer, Integer> tm = new TreeMap<>();
for (int x : nums) {
tm.merge(x, 1, Integer::sum);
}
while (!tm.isEmpty()) {
int x = tm.firstKey();
for (int y = x; y < x + k; ++y) {
int t = tm.merge(y, -1, Integer::sum);
if (t < 0) {
return false;
}
if (t == 0) {
tm.remove(y);
}
}
}
return true;
}
}
C++
class Solution {
public:
bool isPossibleDivide(vector<int>& nums, int k) {
if (nums.size() % k) {
return false;
}
map<int, int> mp;
for (int x : nums) {
++mp[x];
}
while (!mp.empty()) {
int x = mp.begin()->first;
for (int y = x; y < x + k; ++y) {
if (!mp.contains(y)) {
return false;
}
if (--mp[y] == 0) {
mp.erase(y);
}
}
}
return true;
}
};
Go
func isPossibleDivide(nums []int, k int) bool {
if len(nums)%k != 0 {
return false
}
tm := treemap.NewWithIntComparator()
for _, x := range nums {
if v, ok := tm.Get(x); ok {
tm.Put(x, v.(int)+1)
} else {
tm.Put(x, 1)
}
}
for !tm.Empty() {
x, _ := tm.Min()
for y := x.(int); y < x.(int)+k; y++ {
if v, ok := tm.Get(y); ok {
if v.(int) == 1 {
tm.Remove(y)
} else {
tm.Put(y, v.(int)-1)
}
} else {
return false
}
}
}
return true
}
TypeScript
function isPossibleDivide(nums: number[], k: number): boolean {
if (nums.length % k !== 0) {
return false;
}
const tm = new TreeMap<number, number>();
for (const x of nums) {
tm.set(x, (tm.get(x) || 0) + 1);
}
while (tm.size()) {
const x = tm.first()![0];
for (let y = x; y < x + k; ++y) {
if (!tm.has(y)) {
return false;
}
if (tm.get(y)! === 1) {
tm.delete(y);
} else {
tm.set(y, tm.get(y)! - 1);
}
}
}
return true;
}
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;
}
search(predicate: (val: T) => boolean, direction: 'left' | 'right'): T | undefined {
let p = this.root;
let result = null;
while (p) {
if (predicate(p.data)) {
result = p;
p = p[direction];
} else {
p = p[direction === 'left' ? 'right' : 'left'];
}
}
return result?.data;
}
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;
}
count(data: T): number {
const node = this.find(data);
return node ? node.count : 0;
}
*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 TreeMap<K = number, V = unknown> {
_size: number;
tree: RBTree<K>;
map: Map<K, V> = new Map();
compare: Compare<K>;
constructor(
collection: Array<[K, V]> | Compare<K> = [],
compare: Compare<K> = (l: K, r: K) => (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 [key, val] of collection) this.set(key, val);
}
size(): number {
return this._size;
}
has(key: K): boolean {
return !!this.tree.find(key);
}
get(key: K): V | undefined {
return this.map.get(key);
}
set(key: K, val: V): boolean {
const successful = this.tree.insert(key);
this._size += successful ? 1 : 0;
this.map.set(key, val);
return successful;
}
delete(key: K): boolean {
const deleted = this.tree.deleteAll(key);
this._size -= deleted ? 1 : 0;
return deleted;
}
ceil(target: K): [K, V] | undefined {
return this.toKeyValue(this.tree.search(key => this.compare(key, target) >= 0, 'left'));
}
floor(target: K): [K, V] | undefined {
return this.toKeyValue(this.tree.search(key => this.compare(key, target) <= 0, 'right'));
}
higher(target: K): [K, V] | undefined {
return this.toKeyValue(this.tree.search(key => this.compare(key, target) > 0, 'left'));
}
lower(target: K): [K, V] | undefined {
return this.toKeyValue(this.tree.search(key => this.compare(key, target) < 0, 'right'));
}
first(): [K, V] | undefined {
return this.toKeyValue(this.tree.inOrder().next().value);
}
last(): [K, V] | undefined {
return this.toKeyValue(this.tree.reverseInOrder().next().value);
}
shift(): [K, V] | undefined {
const first = this.first();
if (first === undefined) return undefined;
this.delete(first[0]);
return first;
}
pop(): [K, V] | undefined {
const last = this.last();
if (last === undefined) return undefined;
this.delete(last[0]);
return last;
}
toKeyValue(key: K): [K, V];
toKeyValue(key: undefined): undefined;
toKeyValue(key: K | undefined): [K, V] | undefined;
toKeyValue(key: K | undefined): [K, V] | undefined {
return key != null ? [key, this.map.get(key)!] : undefined;
}
*[Symbol.iterator](): Generator<[K, V], void, void> {
for (const key of this.keys()) yield this.toKeyValue(key);
}
*keys(): Generator<K, void, void> {
for (const key of this.tree.inOrder()) yield key;
}
*values(): Generator<V, undefined, void> {
for (const key of this.keys()) yield this.map.get(key)!;
return undefined;
}
*rkeys(): Generator<K, undefined, void> {
for (const key of this.tree.reverseInOrder()) yield key;
return undefined;
}
*rvalues(): Generator<V, undefined, void> {
for (const key of this.rkeys()) yield this.map.get(key)!;
return undefined;
}
}