1438. Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit
Description
Given an array of integers nums and an integer limit, return the size of the longest non-empty subarray such that the absolute difference between any two elements of this subarray is less than or equal to limit.
Example 1:
Input: nums = [8,2,4,7], limit = 4 Output: 2 Explanation: All subarrays are: [8] with maximum absolute diff |8-8| = 0 <= 4. [8,2] with maximum absolute diff |8-2| = 6 > 4. [8,2,4] with maximum absolute diff |8-2| = 6 > 4. [8,2,4,7] with maximum absolute diff |8-2| = 6 > 4. [2] with maximum absolute diff |2-2| = 0 <= 4. [2,4] with maximum absolute diff |2-4| = 2 <= 4. [2,4,7] with maximum absolute diff |2-7| = 5 > 4. [4] with maximum absolute diff |4-4| = 0 <= 4. [4,7] with maximum absolute diff |4-7| = 3 <= 4. [7] with maximum absolute diff |7-7| = 0 <= 4. Therefore, the size of the longest subarray is 2.
Example 2:
Input: nums = [10,1,2,4,7,2], limit = 5 Output: 4 Explanation: The subarray [2,4,7,2] is the longest since the maximum absolute diff is |2-7| = 5 <= 5.
Example 3:
Input: nums = [4,2,2,2,4,4,2,2], limit = 0 Output: 3
Constraints:
1 <= nums.length <= 1051 <= nums[i] <= 1090 <= limit <= 109
Solutions
Solution 1: Ordered Set + Sliding Window
We can enumerate each position as the right endpoint of the subarray, and find the leftmost left endpoint corresponding to it, such that the difference between the maximum and minimum values in the interval does not exceed $limit$. During the process, we use an ordered set to maintain the maximum and minimum values within the window.
The time complexity is $O(n \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array nums.
Python3
class Solution:
def longestSubarray(self, nums: List[int], limit: int) -> int:
sl = SortedList()
ans = j = 0
for i, x in enumerate(nums):
sl.add(x)
while sl[-1] - sl[0] > limit:
sl.remove(nums[j])
j += 1
ans = max(ans, i - j + 1)
return ans
Java
class Solution {
public int longestSubarray(int[] nums, int limit) {
TreeMap<Integer, Integer> tm = new TreeMap<>();
int ans = 0;
for (int i = 0, j = 0; i < nums.length; ++i) {
tm.merge(nums[i], 1, Integer::sum);
for (; tm.lastKey() - tm.firstKey() > limit; ++j) {
if (tm.merge(nums[j], -1, Integer::sum) == 0) {
tm.remove(nums[j]);
}
}
ans = Math.max(ans, i - j + 1);
}
return ans;
}
}
C++
class Solution {
public:
int longestSubarray(vector<int>& nums, int limit) {
multiset<int> s;
int ans = 0, j = 0;
for (int i = 0; i < nums.size(); ++i) {
s.insert(nums[i]);
while (*s.rbegin() - *s.begin() > limit) {
s.erase(s.find(nums[j++]));
}
ans = max(ans, i - j + 1);
}
return ans;
}
};
Go
func longestSubarray(nums []int, limit int) (ans int) {
merge := func(st *redblacktree.Tree[int, int], x, v int) {
c, _ := st.Get(x)
if c+v == 0 {
st.Remove(x)
} else {
st.Put(x, c+v)
}
}
st := redblacktree.New[int, int]()
j := 0
for i, x := range nums {
merge(st, x, 1)
for ; st.Right().Key-st.Left().Key > limit; j++ {
merge(st, nums[j], -1)
}
ans = max(ans, i-j+1)
}
return
}
TypeScript
function longestSubarray(nums: number[], limit: number): number {
const ts = new TreapMultiSet<number>();
let ans = 0;
let j = 0;
for (let i = 0; i < nums.length; ++i) {
ts.add(nums[i]);
while (ts.last()! - ts.first()! > limit) {
ts.delete(nums[j++]);
}
ans = Math.max(ans, i - j + 1);
}
return ans;
}
type CompareFunction<T, R extends 'number' | 'boolean'> = (
a: T,
b: T,
) => R extends 'number' ? number : boolean;
interface ITreapMultiSet<T> extends Iterable<T> {
add: (...value: T[]) => this;
has: (value: T) => boolean;
delete: (value: T) => void;
bisectLeft: (value: T) => number;
bisectRight: (value: T) => number;
indexOf: (value: T) => number;
lastIndexOf: (value: T) => number;
at: (index: number) => T | undefined;
first: () => T | undefined;
last: () => T | undefined;
lower: (value: T) => T | undefined;
higher: (value: T) => T | undefined;
floor: (value: T) => T | undefined;
ceil: (value: T) => T | undefined;
shift: () => T | undefined;
pop: (index?: number) => T | undefined;
count: (value: T) => number;
keys: () => IterableIterator<T>;
values: () => IterableIterator<T>;
rvalues: () => IterableIterator<T>;
entries: () => IterableIterator<[number, T]>;
readonly size: number;
}
class TreapNode<T = number> {
value: T;
count: number;
size: number;
priority: number;
left: TreapNode<T> | null;
right: TreapNode<T> | null;
constructor(value: T) {
this.value = value;
this.count = 1;
this.size = 1;
this.priority = Math.random();
this.left = null;
this.right = null;
}
static getSize(node: TreapNode<any> | null): number {
return node?.size ?? 0;
}
static getFac(node: TreapNode<any> | null): number {
return node?.priority ?? 0;
}
pushUp(): void {
let tmp = this.count;
tmp += TreapNode.getSize(this.left);
tmp += TreapNode.getSize(this.right);
this.size = tmp;
}
rotateRight(): TreapNode<T> {
// eslint-disable-next-line @typescript-eslint/no-this-alias
let node: TreapNode<T> = this;
const left = node.left;
node.left = left?.right ?? null;
left && (left.right = node);
left && (node = left);
node.right?.pushUp();
node.pushUp();
return node;
}
rotateLeft(): TreapNode<T> {
// eslint-disable-next-line @typescript-eslint/no-this-alias
let node: TreapNode<T> = this;
const right = node.right;
node.right = right?.left ?? null;
right && (right.left = node);
right && (node = right);
node.left?.pushUp();
node.pushUp();
return node;
}
}
class TreapMultiSet<T = number> implements ITreapMultiSet<T> {
private readonly root: TreapNode<T>;
private readonly compareFn: CompareFunction<T, 'number'>;
private readonly leftBound: T;
private readonly rightBound: T;
constructor(compareFn?: CompareFunction<T, 'number'>);
constructor(compareFn: CompareFunction<T, 'number'>, leftBound: T, rightBound: T);
constructor(
compareFn: CompareFunction<T, any> = (a: any, b: any) => a - b,
leftBound: any = -Infinity,
rightBound: any = Infinity,
) {
this.root = new TreapNode<T>(rightBound);
this.root.priority = Infinity;
this.root.left = new TreapNode<T>(leftBound);
this.root.left.priority = -Infinity;
this.root.pushUp();
this.leftBound = leftBound;
this.rightBound = rightBound;
this.compareFn = compareFn;
}
get size(): number {
return this.root.size - 2;
}
get height(): number {
const getHeight = (node: TreapNode<T> | null): number => {
if (node == null) return 0;
return 1 + Math.max(getHeight(node.left), getHeight(node.right));
};
return getHeight(this.root);
}
/**
*
* @complexity `O(logn)`
* @description Returns true if value is a member.
*/
has(value: T): boolean {
const compare = this.compareFn;
const dfs = (node: TreapNode<T> | null, value: T): boolean => {
if (node == null) return false;
if (compare(node.value, value) === 0) return true;
if (compare(node.value, value) < 0) return dfs(node.right, value);
return dfs(node.left, value);
};
return dfs(this.root, value);
}
/**
*
* @complexity `O(logn)`
* @description Add value to sorted set.
*/
add(...values: T[]): this {
const compare = this.compareFn;
const dfs = (
node: TreapNode<T> | null,
value: T,
parent: TreapNode<T>,
direction: 'left' | 'right',
): void => {
if (node == null) return;
if (compare(node.value, value) === 0) {
node.count++;
node.pushUp();
} else if (compare(node.value, value) > 0) {
if (node.left) {
dfs(node.left, value, node, 'left');
} else {
node.left = new TreapNode(value);
node.pushUp();
}
if (TreapNode.getFac(node.left) > node.priority) {
parent[direction] = node.rotateRight();
}
} else if (compare(node.value, value) < 0) {
if (node.right) {
dfs(node.right, value, node, 'right');
} else {
node.right = new TreapNode(value);
node.pushUp();
}
if (TreapNode.getFac(node.right) > node.priority) {
parent[direction] = node.rotateLeft();
}
}
parent.pushUp();
};
values.forEach(value => dfs(this.root.left, value, this.root, 'left'));
return this;
}
/**
*
* @complexity `O(logn)`
* @description Remove value from sorted set if it is a member.
* If value is not a member, do nothing.
*/
delete(value: T): void {
const compare = this.compareFn;
const dfs = (
node: TreapNode<T> | null,
value: T,
parent: TreapNode<T>,
direction: 'left' | 'right',
): void => {
if (node == null) return;
if (compare(node.value, value) === 0) {
if (node.count > 1) {
node.count--;
node?.pushUp();
} else if (node.left == null && node.right == null) {
parent[direction] = null;
} else {
// 旋到根节点
if (
node.right == null ||
TreapNode.getFac(node.left) > TreapNode.getFac(node.right)
) {
parent[direction] = node.rotateRight();
dfs(parent[direction]?.right ?? null, value, parent[direction]!, 'right');
} else {
parent[direction] = node.rotateLeft();
dfs(parent[direction]?.left ?? null, value, parent[direction]!, 'left');
}
}
} else if (compare(node.value, value) > 0) {
dfs(node.left, value, node, 'left');
} else if (compare(node.value, value) < 0) {
dfs(node.right, value, node, 'right');
}
parent?.pushUp();
};
dfs(this.root.left, value, this.root, 'left');
}
/**
*
* @complexity `O(logn)`
* @description Returns an index to insert value in the sorted set.
* If the value is already present, the insertion point will be before (to the left of) any existing values.
*/
bisectLeft(value: T): number {
const compare = this.compareFn;
const dfs = (node: TreapNode<T> | null, value: T): number => {
if (node == null) return 0;
if (compare(node.value, value) === 0) {
return TreapNode.getSize(node.left);
} else if (compare(node.value, value) > 0) {
return dfs(node.left, value);
} else if (compare(node.value, value) < 0) {
return dfs(node.right, value) + TreapNode.getSize(node.left) + node.count;
}
return 0;
};
return dfs(this.root, value) - 1;
}
/**
*
* @complexity `O(logn)`
* @description Returns an index to insert value in the sorted set.
* If the value is already present, the insertion point will be before (to the right of) any existing values.
*/
bisectRight(value: T): number {
const compare = this.compareFn;
const dfs = (node: TreapNode<T> | null, value: T): number => {
if (node == null) return 0;
if (compare(node.value, value) === 0) {
return TreapNode.getSize(node.left) + node.count;
} else if (compare(node.value, value) > 0) {
return dfs(node.left, value);
} else if (compare(node.value, value) < 0) {
return dfs(node.right, value) + TreapNode.getSize(node.left) + node.count;
}
return 0;
};
return dfs(this.root, value) - 1;
}
/**
*
* @complexity `O(logn)`
* @description Returns the index of the first occurrence of a value in the set, or -1 if it is not present.
*/
indexOf(value: T): number {
const compare = this.compareFn;
let isExist = false;
const dfs = (node: TreapNode<T> | null, value: T): number => {
if (node == null) return 0;
if (compare(node.value, value) === 0) {
isExist = true;
return TreapNode.getSize(node.left);
} else if (compare(node.value, value) > 0) {
return dfs(node.left, value);
} else if (compare(node.value, value) < 0) {
return dfs(node.right, value) + TreapNode.getSize(node.left) + node.count;
}
return 0;
};
const res = dfs(this.root, value) - 1;
return isExist ? res : -1;
}
/**
*
* @complexity `O(logn)`
* @description Returns the index of the last occurrence of a value in the set, or -1 if it is not present.
*/
lastIndexOf(value: T): number {
const compare = this.compareFn;
let isExist = false;
const dfs = (node: TreapNode<T> | null, value: T): number => {
if (node == null) return 0;
if (compare(node.value, value) === 0) {
isExist = true;
return TreapNode.getSize(node.left) + node.count - 1;
} else if (compare(node.value, value) > 0) {
return dfs(node.left, value);
} else if (compare(node.value, value) < 0) {
return dfs(node.right, value) + TreapNode.getSize(node.left) + node.count;
}
return 0;
};
const res = dfs(this.root, value) - 1;
return isExist ? res : -1;
}
/**
*
* @complexity `O(logn)`
* @description Returns the item located at the specified index.
* @param index The zero-based index of the desired code unit. A negative index will count back from the last item.
*/
at(index: number): T | undefined {
if (index < 0) index += this.size;
if (index < 0 || index >= this.size) return undefined;
const dfs = (node: TreapNode<T> | null, rank: number): T | undefined => {
if (node == null) return undefined;
if (TreapNode.getSize(node.left) >= rank) {
return dfs(node.left, rank);
} else if (TreapNode.getSize(node.left) + node.count >= rank) {
return node.value;
} else {
return dfs(node.right, rank - TreapNode.getSize(node.left) - node.count);
}
};
const res = dfs(this.root, index + 2);
return ([this.leftBound, this.rightBound] as any[]).includes(res) ? undefined : res;
}
/**
*
* @complexity `O(logn)`
* @description Find and return the element less than `val`, return `undefined` if no such element found.
*/
lower(value: T): T | undefined {
const compare = this.compareFn;
const dfs = (node: TreapNode<T> | null, value: T): T | undefined => {
if (node == null) return undefined;
if (compare(node.value, value) >= 0) return dfs(node.left, value);
const tmp = dfs(node.right, value);
if (tmp == null || compare(node.value, tmp) > 0) {
return node.value;
} else {
return tmp;
}
};
const res = dfs(this.root, value) as any;
return res === this.leftBound ? undefined : res;
}
/**
*
* @complexity `O(logn)`
* @description Find and return the element greater than `val`, return `undefined` if no such element found.
*/
higher(value: T): T | undefined {
const compare = this.compareFn;
const dfs = (node: TreapNode<T> | null, value: T): T | undefined => {
if (node == null) return undefined;
if (compare(node.value, value) <= 0) return dfs(node.right, value);
const tmp = dfs(node.left, value);
if (tmp == null || compare(node.value, tmp) < 0) {
return node.value;
} else {
return tmp;
}
};
const res = dfs(this.root, value) as any;
return res === this.rightBound ? undefined : res;
}
/**
*
* @complexity `O(logn)`
* @description Find and return the element less than or equal to `val`, return `undefined` if no such element found.
*/
floor(value: T): T | undefined {
const compare = this.compareFn;
const dfs = (node: TreapNode<T> | null, value: T): T | undefined => {
if (node == null) return undefined;
if (compare(node.value, value) === 0) return node.value;
if (compare(node.value, value) >= 0) return dfs(node.left, value);
const tmp = dfs(node.right, value);
if (tmp == null || compare(node.value, tmp) > 0) {
return node.value;
} else {
return tmp;
}
};
const res = dfs(this.root, value) as any;
return res === this.leftBound ? undefined : res;
}
/**
*
* @complexity `O(logn)`
* @description Find and return the element greater than or equal to `val`, return `undefined` if no such element found.
*/
ceil(value: T): T | undefined {
const compare = this.compareFn;
const dfs = (node: TreapNode<T> | null, value: T): T | undefined => {
if (node == null) return undefined;
if (compare(node.value, value) === 0) return node.value;
if (compare(node.value, value) <= 0) return dfs(node.right, value);
const tmp = dfs(node.left, value);
if (tmp == null || compare(node.value, tmp) < 0) {
return node.value;
} else {
return tmp;
}
};
const res = dfs(this.root, value) as any;
return res === this.rightBound ? undefined : res;
}
/**
* @complexity `O(logn)`
* @description
* Returns the last element from set.
* If the set is empty, undefined is returned.
*/
first(): T | undefined {
const iter = this.inOrder();
iter.next();
const res = iter.next().value;
return res === this.rightBound ? undefined : res;
}
/**
* @complexity `O(logn)`
* @description
* Returns the last element from set.
* If the set is empty, undefined is returned .
*/
last(): T | undefined {
const iter = this.reverseInOrder();
iter.next();
const res = iter.next().value;
return res === this.leftBound ? undefined : res;
}
/**
* @complexity `O(logn)`
* @description
* Removes the first element from an set and returns it.
* If the set is empty, undefined is returned and the set is not modified.
*/
shift(): T | undefined {
const first = this.first();
if (first === undefined) return undefined;
this.delete(first);
return first;
}
/**
* @complexity `O(logn)`
* @description
* Removes the last element from an set and returns it.
* If the set is empty, undefined is returned and the set is not modified.
*/
pop(index?: number): T | undefined {
if (index == null) {
const last = this.last();
if (last === undefined) return undefined;
this.delete(last);
return last;
}
const toDelete = this.at(index);
if (toDelete == null) return;
this.delete(toDelete);
return toDelete;
}
/**
*
* @complexity `O(logn)`
* @description
* Returns number of occurrences of value in the sorted set.
*/
count(value: T): number {
const compare = this.compareFn;
const dfs = (node: TreapNode<T> | null, value: T): number => {
if (node == null) return 0;
if (compare(node.value, value) === 0) return node.count;
if (compare(node.value, value) < 0) return dfs(node.right, value);
return dfs(node.left, value);
};
return dfs(this.root, value);
}
*[Symbol.iterator](): Generator<T, any, any> {
yield* this.values();
}
/**
* @description
* Returns an iterable of keys in the set.
*/
*keys(): Generator<T, any, any> {
yield* this.values();
}
/**
* @description
* Returns an iterable of values in the set.
*/
*values(): Generator<T, any, any> {
const iter = this.inOrder();
iter.next();
const steps = this.size;
for (let _ = 0; _ < steps; _++) {
yield iter.next().value;
}
}
/**
* @description
* Returns a generator for reversed order traversing the set.
*/
*rvalues(): Generator<T, any, any> {
const iter = this.reverseInOrder();
iter.next();
const steps = this.size;
for (let _ = 0; _ < steps; _++) {
yield iter.next().value;
}
}
/**
* @description
* Returns an iterable of key, value pairs for every entry in the set.
*/
*entries(): IterableIterator<[number, T]> {
const iter = this.inOrder();
iter.next();
const steps = this.size;
for (let i = 0; i < steps; i++) {
yield [i, iter.next().value];
}
}
private *inOrder(root: TreapNode<T> | null = this.root): Generator<T, any, any> {
if (root == null) return;
yield* this.inOrder(root.left);
const count = root.count;
for (let _ = 0; _ < count; _++) {
yield root.value;
}
yield* this.inOrder(root.right);
}
private *reverseInOrder(root: TreapNode<T> | null = this.root): Generator<T, any, any> {
if (root == null) return;
yield* this.reverseInOrder(root.right);
const count = root.count;
for (let _ = 0; _ < count; _++) {
yield root.value;
}
yield* this.reverseInOrder(root.left);
}
}
Solution 2: Binary Search + Sliding Window
We notice that if a subarray of length $k$ satisfies the condition, then a subarray of length $k' < k$ also satisfies the condition. This shows a monotonicity, therefore, we can use binary search to find the longest subarray that satisfies the condition.
We define the left boundary of the binary search as $l = 0$, and the right boundary as $r = n$. For each $mid = \frac{l + r + 1}{2}$, we check whether there exists a subarray of length $mid$ that satisfies the condition. If it exists, we update $l = mid$, otherwise we update $r = mid - 1$. The problem is transformed into whether there exists a subarray of length $mid$ in the array that satisfies the condition, which is actually to find the difference between the maximum and minimum values in the sliding window does not exceed $limit$. We can use two monotonic queues to maintain the maximum and minimum values in the window respectively.
The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array $nums$.
Python3
class Solution:
def longestSubarray(self, nums: List[int], limit: int) -> int:
def check(k: int) -> bool:
min_q = deque()
max_q = deque()
for i, x in enumerate(nums):
if min_q and i - min_q[0] + 1 > k:
min_q.popleft()
if max_q and i - max_q[0] + 1 > k:
max_q.popleft()
while min_q and nums[min_q[-1]] >= x:
min_q.pop()
while max_q and nums[max_q[-1]] <= x:
max_q.pop()
min_q.append(i)
max_q.append(i)
if i >= k - 1 and nums[max_q[0]] - nums[min_q[0]] <= limit:
return True
return False
l, r = 1, len(nums)
while l < r:
mid = (l + r + 1) >> 1
if check(mid):
l = mid
else:
r = mid - 1
return l
Java
class Solution {
private int[] nums;
private int limit;
public int longestSubarray(int[] nums, int limit) {
this.nums = nums;
this.limit = limit;
int l = 1, r = nums.length;
while (l < r) {
int mid = (l + r + 1) >> 1;
if (check(mid)) {
l = mid;
} else {
r = mid - 1;
}
}
return l;
}
private boolean check(int k) {
Deque<Integer> minQ = new ArrayDeque<>();
Deque<Integer> maxQ = new ArrayDeque<>();
for (int i = 0; i < nums.length; ++i) {
if (!minQ.isEmpty() && i - minQ.peekFirst() + 1 > k) {
minQ.pollFirst();
}
if (!maxQ.isEmpty() && i - maxQ.peekFirst() + 1 > k) {
maxQ.pollFirst();
}
while (!minQ.isEmpty() && nums[minQ.peekLast()] >= nums[i]) {
minQ.pollLast();
}
while (!maxQ.isEmpty() && nums[maxQ.peekLast()] <= nums[i]) {
maxQ.pollLast();
}
minQ.offer(i);
maxQ.offer(i);
if (i >= k - 1 && nums[maxQ.peekFirst()] - nums[minQ.peekFirst()] <= limit) {
return true;
}
}
return false;
}
}
C++
class Solution {
public:
int longestSubarray(vector<int>& nums, int limit) {
auto check = [&](int k) {
deque<int> min_q;
deque<int> max_q;
for (int i = 0; i < nums.size(); ++i) {
if (!min_q.empty() && i - min_q.front() + 1 > k) {
min_q.pop_front();
}
if (!max_q.empty() && i - max_q.front() + 1 > k) {
max_q.pop_front();
}
while (!min_q.empty() && nums[min_q.back()] >= nums[i]) {
min_q.pop_back();
}
while (!max_q.empty() && nums[max_q.back()] <= nums[i]) {
max_q.pop_back();
}
min_q.push_back(i);
max_q.push_back(i);
if (i >= k - 1 && nums[max_q.front()] - nums[min_q.front()] <= limit) {
return true;
}
}
return false;
};
int l = 1, r = nums.size();
while (l < r) {
int mid = (l + r + 1) >> 1;
if (check(mid)) {
l = mid;
} else {
r = mid - 1;
}
}
return l;
}
};
Go
func longestSubarray(nums []int, limit int) int {
l, r := 0, len(nums)
check := func(k int) bool {
minq := Deque{}
maxq := Deque{}
for i, x := range nums {
for !minq.Empty() && i-minq.Front()+1 > k {
minq.PopFront()
}
for !maxq.Empty() && i-maxq.Front()+1 > k {
maxq.PopFront()
}
for !minq.Empty() && nums[minq.Back()] >= x {
minq.PopBack()
}
for !maxq.Empty() && nums[maxq.Back()] <= x {
maxq.PopBack()
}
minq.PushBack(i)
maxq.PushBack(i)
if i >= k-1 && nums[maxq.Front()]-nums[minq.Front()] <= limit {
return true
}
}
return false
}
for l < r {
mid := (l + r + 1) >> 1
if check(mid) {
l = mid
} else {
r = mid - 1
}
}
return l
}
// template
type Deque struct{ l, r []int }
func (q Deque) Empty() bool {
return len(q.l) == 0 && len(q.r) == 0
}
func (q Deque) Size() int {
return len(q.l) + len(q.r)
}
func (q *Deque) PushFront(v int) {
q.l = append(q.l, v)
}
func (q *Deque) PushBack(v int) {
q.r = append(q.r, v)
}
func (q *Deque) PopFront() (v int) {
if len(q.l) > 0 {
q.l, v = q.l[:len(q.l)-1], q.l[len(q.l)-1]
} else {
v, q.r = q.r[0], q.r[1:]
}
return
}
func (q *Deque) PopBack() (v int) {
if len(q.r) > 0 {
q.r, v = q.r[:len(q.r)-1], q.r[len(q.r)-1]
} else {
v, q.l = q.l[0], q.l[1:]
}
return
}
func (q Deque) Front() int {
if len(q.l) > 0 {
return q.l[len(q.l)-1]
}
return q.r[0]
}
func (q Deque) Back() int {
if len(q.r) > 0 {
return q.r[len(q.r)-1]
}
return q.l[0]
}
func (q Deque) Get(i int) int {
if i < len(q.l) {
return q.l[len(q.l)-1-i]
}
return q.r[i-len(q.l)]
}
TypeScript
function longestSubarray(nums: number[], limit: number): number {
const n = nums.length;
let [l, r] = [0, n];
const check = (k: number): boolean => {
const minq = new Deque<number>();
const maxq = new Deque<number>();
for (let i = 0; i < n; ++i) {
while (!minq.isEmpty() && i - minq.frontValue()! + 1 > k) {
minq.popFront();
}
while (!maxq.isEmpty() && i - maxq.frontValue()! + 1 > k) {
maxq.popFront();
}
while (!minq.isEmpty() && nums[minq.backValue()!] >= nums[i]) {
minq.popBack();
}
while (!maxq.isEmpty() && nums[maxq.backValue()!] <= nums[i]) {
maxq.popBack();
}
minq.pushBack(i);
maxq.pushBack(i);
if (i >= k - 1 && nums[maxq.frontValue()!] - nums[minq.frontValue()!] <= limit) {
return true;
}
}
return false;
};
while (l < r) {
const mid = (l + r + 1) >> 1;
if (check(mid)) {
l = mid;
} else {
r = mid - 1;
}
}
return l;
}
class Node<T> {
value: T;
next: Node<T> | null;
prev: Node<T> | null;
constructor(value: T) {
this.value = value;
this.next = null;
this.prev = null;
}
}
class Deque<T> {
private front: Node<T> | null;
private back: Node<T> | null;
private size: number;
constructor() {
this.front = null;
this.back = null;
this.size = 0;
}
pushFront(val: T): void {
const newNode = new Node(val);
if (this.isEmpty()) {
this.front = newNode;
this.back = newNode;
} else {
newNode.next = this.front;
this.front!.prev = newNode;
this.front = newNode;
}
this.size++;
}
pushBack(val: T): void {
const newNode = new Node(val);
if (this.isEmpty()) {
this.front = newNode;
this.back = newNode;
} else {
newNode.prev = this.back;
this.back!.next = newNode;
this.back = newNode;
}
this.size++;
}
popFront(): T | undefined {
if (this.isEmpty()) {
return undefined;
}
const value = this.front!.value;
this.front = this.front!.next;
if (this.front !== null) {
this.front.prev = null;
} else {
this.back = null;
}
this.size--;
return value;
}
popBack(): T | undefined {
if (this.isEmpty()) {
return undefined;
}
const value = this.back!.value;
this.back = this.back!.prev;
if (this.back !== null) {
this.back.next = null;
} else {
this.front = null;
}
this.size--;
return value;
}
frontValue(): T | undefined {
return this.front?.value;
}
backValue(): T | undefined {
return this.back?.value;
}
getSize(): number {
return this.size;
}
isEmpty(): boolean {
return this.size === 0;
}
}
Solution 3: Sliding Window + Deque
We can use a deque to maintain the maximum and minimum values within the window. We maintain two deques, one for storing the indices of the maximum values and the other for the minimum values within the window. Define two pointers $l$ and $r$ to point to the left and right boundaries of the window, respectively.
Each time we move the right boundary $r$ to the right, we check if the element corresponding to the tail index of the maximum value deque is less than the current element. If it is, we dequeue the tail element until the element corresponding to the tail of the maximum value deque is not less than the current element. Similarly, we check if the element corresponding to the tail index of the minimum value deque is greater than the current element. If it is, we dequeue the tail element until the element corresponding to the tail of the minimum value deque is not greater than the current element. Then, we enqueue the current element's index.
If the difference between the elements at the front of the maximum value deque and the minimum value deque is greater than $limit$, then we move the left boundary $l$ to the right. If the element at the front of the maximum value deque is less than $l$, we dequeue the front element of the maximum value deque. Similarly, if the element at the front of the minimum value deque is less than $l$, we dequeue the front element of the minimum value deque.
The answer is $n - l$.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.
Python3
class Solution:
def longestSubarray(self, nums: List[int], limit: int) -> int:
maxq = deque()
minq = deque()
l, n = 0, len(nums)
for r, x in enumerate(nums):
while maxq and nums[maxq[-1]] < x:
maxq.pop()
while minq and nums[minq[-1]] > x:
minq.pop()
maxq.append(r)
minq.append(r)
if nums[maxq[0]] - nums[minq[0]] > limit:
l += 1
if maxq[0] < l:
maxq.popleft()
if minq[0] < l:
minq.popleft()
return n - l
Java
class Solution {
public int longestSubarray(int[] nums, int limit) {
Deque<Integer> maxQ = new ArrayDeque<>();
Deque<Integer> minQ = new ArrayDeque<>();
int n = nums.length;
int l = 0;
for (int r = 0; r < n; ++r) {
while (!maxQ.isEmpty() && nums[maxQ.peekLast()] < nums[r]) {
maxQ.pollLast();
}
while (!minQ.isEmpty() && nums[minQ.peekLast()] > nums[r]) {
minQ.pollLast();
}
maxQ.offerLast(r);
minQ.offerLast(r);
if (nums[maxQ.peekFirst()] - nums[minQ.peekFirst()] > limit) {
++l;
if (maxQ.peekFirst() < l) {
maxQ.pollFirst();
}
if (minQ.peekFirst() < l) {
minQ.pollFirst();
}
}
}
return n - l;
}
}
C++
class Solution {
public:
int longestSubarray(vector<int>& nums, int limit) {
deque<int> max_q;
deque<int> min_q;
int n = nums.size();
int l = 0;
for (int r = 0; r < n; ++r) {
while (!max_q.empty() && nums[max_q.back()] < nums[r]) {
max_q.pop_back();
}
while (!min_q.empty() && nums[min_q.back()] > nums[r]) {
min_q.pop_back();
}
max_q.push_back(r);
min_q.push_back(r);
if (nums[max_q.front()] - nums[min_q.front()] > limit) {
++l;
if (max_q.front() < l) {
max_q.pop_front();
}
if (min_q.front() < l) {
min_q.pop_front();
}
}
}
return n - l;
}
};
Go
func longestSubarray(nums []int, limit int) int {
var maxq, minq Deque
n := len(nums)
l := 0
for r, x := range nums {
for !maxq.Empty() && nums[maxq.Back()] < x {
maxq.PopBack()
}
for !minq.Empty() && nums[minq.Back()] > x {
minq.PopBack()
}
maxq.PushBack(r)
minq.PushBack(r)
if nums[maxq.Front()]-nums[minq.Front()] > limit {
l++
if maxq.Front() < l {
maxq.PopFront()
}
if minq.Front() < l {
minq.PopFront()
}
}
}
return n - l
}
type Deque struct{ l, r []int }
func (q Deque) Empty() bool {
return len(q.l) == 0 && len(q.r) == 0
}
func (q Deque) Size() int {
return len(q.l) + len(q.r)
}
func (q *Deque) PushFront(v int) {
q.l = append(q.l, v)
}
func (q *Deque) PushBack(v int) {
q.r = append(q.r, v)
}
func (q *Deque) PopFront() (v int) {
if len(q.l) > 0 {
q.l, v = q.l[:len(q.l)-1], q.l[len(q.l)-1]
} else {
v, q.r = q.r[0], q.r[1:]
}
return
}
func (q *Deque) PopBack() (v int) {
if len(q.r) > 0 {
q.r, v = q.r[:len(q.r)-1], q.r[len(q.r)-1]
} else {
v, q.l = q.l[0], q.l[1:]
}
return
}
func (q Deque) Front() int {
if len(q.l) > 0 {
return q.l[len(q.l)-1]
}
return q.r[0]
}
func (q Deque) Back() int {
if len(q.r) > 0 {
return q.r[len(q.r)-1]
}
return q.l[0]
}
func (q Deque) Get(i int) int {
if i < len(q.l) {
return q.l[len(q.l)-1-i]
}
return q.r[i-len(q.l)]
}
TypeScript
function longestSubarray(nums: number[], limit: number): number {
const n = nums.length;
let [h1, t1] = [0, -1];
let [h2, t2] = [0, -1];
let l = 0;
const maxq = Array(n);
const minq = Array(n);
for (let r = 0; r < n; ++r) {
while (h1 <= t1 && nums[maxq[t1]] < nums[r]) {
--t1;
}
while (h2 <= t2 && nums[minq[t2]] > nums[r]) {
--t2;
}
maxq[++t1] = r;
minq[++t2] = r;
if (nums[maxq[h1]] - nums[minq[h2]] > limit) {
++l;
if (maxq[h1] < l) {
++h1;
}
if (minq[h2] < l) {
++h2;
}
}
}
return n - l;
}