846. Hand of Straights
Description
Alice has some number of cards and she wants to rearrange the cards into groups so that each group is of size groupSize, and consists of groupSize consecutive cards.
Given an integer array hand where hand[i] is the value written on the ith card and an integer groupSize, return true if she can rearrange the cards, or false otherwise.
Example 1:
Input: hand = [1,2,3,6,2,3,4,7,8], groupSize = 3 Output: true Explanation: Alice's hand can be rearranged as [1,2,3],[2,3,4],[6,7,8]
Example 2:
Input: hand = [1,2,3,4,5], groupSize = 4 Output: false Explanation: Alice's hand can not be rearranged into groups of 4.
Constraints:
1 <= hand.length <= 1040 <= hand[i] <= 1091 <= groupSize <= hand.length
Note: This question is the same as 1296: https://leetcode.com/problems/divide-array-in-sets-of-k-consecutive-numbers/
Solutions
Solution 1: Hash Table + Sorting
We first check whether the length of the array $\textit{hand}$ is divisible by $\textit{groupSize}$. If it is not, this means that the array cannot be partitioned into multiple subarrays of length $\textit{groupSize}$, so we return $\text{false}$.
Next, we use a hash table $\textit{cnt}$ to count the occurrences of each number in the array $\textit{hand}$, and then we sort the array $\textit{hand}$.
After that, we iterate over the sorted array $\textit{hand}$. For each number $x$, if $\textit{cnt}[x] \neq 0$, we enumerate every number $y$ from $x$ to $x + \textit{groupSize} - 1$. If $\textit{cnt}[y] = 0$, it means that we cannot partition the array into multiple subarrays of length $\textit{groupSize}$, so we return $\text{false}$. Otherwise, we decrement $\textit{cnt}[y]$ by $1$.
If the iteration completes successfully, it means that the array can be partitioned into multiple valid subarrays, so we return $\text{true}$.
The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$, where $n$ is the length of the array $\textit{hand}$.
Python3
class Solution:
def isNStraightHand(self, hand: List[int], groupSize: int) -> bool:
if len(hand) % groupSize:
return False
cnt = Counter(hand)
for x in sorted(hand):
if cnt[x]:
for y in range(x, x + groupSize):
if cnt[y] == 0:
return False
cnt[y] -= 1
return True
Java
class Solution {
public boolean isNStraightHand(int[] hand, int groupSize) {
if (hand.length % groupSize != 0) {
return false;
}
Arrays.sort(hand);
Map<Integer, Integer> cnt = new HashMap<>();
for (int x : hand) {
cnt.merge(x, 1, Integer::sum);
}
for (int x : hand) {
if (cnt.getOrDefault(x, 0) > 0) {
for (int y = x; y < x + groupSize; ++y) {
if (cnt.merge(y, -1, Integer::sum) < 0) {
return false;
}
}
}
}
return true;
}
}
C++
class Solution {
public:
bool isNStraightHand(vector<int>& hand, int groupSize) {
if (hand.size() % groupSize) {
return false;
}
ranges::sort(hand);
unordered_map<int, int> cnt;
for (int x : hand) {
++cnt[x];
}
for (int x : hand) {
if (cnt.contains(x)) {
for (int y = x; y < x + groupSize; ++y) {
if (!cnt.contains(y)) {
return false;
}
if (--cnt[y] == 0) {
cnt.erase(y);
}
}
}
}
return true;
}
};
Go
func isNStraightHand(hand []int, groupSize int) bool {
if len(hand)%groupSize != 0 {
return false
}
sort.Ints(hand)
cnt := map[int]int{}
for _, x := range hand {
cnt[x]++
}
for _, x := range hand {
if cnt[x] > 0 {
for y := x; y < x+groupSize; y++ {
if cnt[y] == 0 {
return false
}
cnt[y]--
}
}
}
return true
}
TypeScript
function isNStraightHand(hand: number[], groupSize: number): boolean {
if (hand.length % groupSize !== 0) {
return false;
}
const cnt = new Map<number, number>();
for (const x of hand) {
cnt.set(x, (cnt.get(x) || 0) + 1);
}
hand.sort((a, b) => a - b);
for (const x of hand) {
if (cnt.get(x)! > 0) {
for (let y = x; y < x + groupSize; 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 whether the length of the array $\textit{hand}$ is divisible by $\textit{groupSize}$. If it is not, this means that the array cannot be partitioned into multiple subarrays of length $\textit{groupSize}$, so we return $\text{false}$.
Next, we use an ordered set $\textit{sd}$ to count the occurrences of each number in the array $\textit{hand}$.
Then, we repeatedly take the smallest value $x$ from the ordered set and enumerate each number $y$ from $x$ to $x + \textit{groupSize} - 1$. If all these numbers appear at least once in the ordered set, we decrement their occurrence count by $1$. If any count reaches $0$, we remove that number from the ordered set. Otherwise, if we encounter a number that does not exist in the ordered set, it means that the array cannot be partitioned into valid subarrays, so we return $\text{false}$. If the iteration completes successfully, it means that the array can be partitioned into multiple valid subarrays, so we return $\text{true}$.
The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$, where $n$ is the length of the array $\textit{hand}$.
Python3
class Solution:
def isNStraightHand(self, hand: List[int], groupSize: int) -> bool:
if len(hand) % groupSize:
return False
cnt = Counter(hand)
sd = SortedDict(cnt)
while sd:
x = next(iter(sd))
for y in range(x, x + groupSize):
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 isNStraightHand(int[] hand, int groupSize) {
if (hand.length % groupSize != 0) {
return false;
}
TreeMap<Integer, Integer> tm = new TreeMap<>();
for (int x : hand) {
tm.merge(x, 1, Integer::sum);
}
while (!tm.isEmpty()) {
int x = tm.firstKey();
for (int y = x; y < x + groupSize; ++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 isNStraightHand(vector<int>& hand, int groupSize) {
if (hand.size() % groupSize) {
return false;
}
map<int, int> mp;
for (int x : hand) {
++mp[x];
}
while (!mp.empty()) {
int x = mp.begin()->first;
for (int y = x; y < x + groupSize; ++y) {
if (!mp.contains(y)) {
return false;
}
if (--mp[y] == 0) {
mp.erase(y);
}
}
}
return true;
}
};
Go
func isNStraightHand(hand []int, groupSize int) bool {
if len(hand)%groupSize != 0 {
return false
}
tm := treemap.NewWithIntComparator()
for _, x := range hand {
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)+groupSize; 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 isNStraightHand(hand: number[], groupSize: number): boolean {
if (hand.length % groupSize !== 0) {
return false;
}
const tm = new TreeMap<number, number>();
for (const x of hand) {
tm.set(x, (tm.get(x) || 0) + 1);
}
while (tm.size()) {
const x = tm.first()![0];
for (let y = x; y < x + groupSize; ++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;
}
}