Formatted question description: https://leetcode.ca/all/1438.html

1438. Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit (Medium)

Given an array of integers nums and an integer limit, return the size of the longest continuous subarray such that the absolute difference between any two elements is less than or equal to limit.

In case there is no subarray satisfying the given condition return 0.

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 <= 10^5
1 <= nums[i] <= 10^9
0 <= limit <= 10^9

Related Topics:
Array, Sliding Window

Solution 1. Sliding Window

Use a multiset<int> s to keep track of the values within the window. We keep extending the right-hand side of the window indexed by j and add A[j] into s.

Whenever we find that the difference between max value and min value in s is greater than k, we shrink the window by moving the left-hand side of the window indexed by i. For every A[i] that we popped out from the window, we erase it from s as well.

In this way, we make sure the window is always valid. The maximum j - i is the answer.

// OJ: https://leetcode.com/problems/longest-continuous-subarray-with-absolute-diff-less-than-or-equal-to-limit/
// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
    int longestSubarray(vector<int>& A, int k) {
        multiset<int> s{ A[0] };
        int i = 0, j = 1, N = A.size(), ans = 1;
        while (j < N) {
            s.insert(A[j++]);
            while (*s.rbegin() - *s.begin() > k) s.erase(s.find(A[i++]));
            ans = max(ans, j - i);
        }
        return ans;
    }
};

Solution 2. Monotonous Deque

We use monotonous deques to keep track of the max/min values within the window. See 239. Sliding Window Maximum (Hard) for the idea.

// OJ: https://leetcode.com/problems/longest-continuous-subarray-with-absolute-diff-less-than-or-equal-to-limit/
// Time: O(N)
// Space: O(N)
class Solution {
public:
    int longestSubarray(vector<int>& A, int k) {
        deque<int> mx{A[0]}, mn{A[0]};
        int N = A.size(), ans = 1, i = 0, j = 1;
        while (j < N) {
            while (j < N && mx.front() - mn.front() <= k) { 
                while (mx.size() && mx.back() < A[j]) mx.pop_back();
                mx.push_back(A[j]);
                while (mn.size() && mn.back() > A[j]) mn.pop_back();
                mn.push_back(A[j]);
                ++j;
                if (mx.front() - mn.front() <= k) ans = max(ans, j - i);
            }
            while (i < j && mx.front() - mn.front() > k) {
                if (A[i] == mx.front()) mx.pop_front();
                if (A[i] == mn.front()) mn.pop_front();
                ++i;
            }
        }
        return ans;
    }
};

Solution 3. Monotonous Deque

Similar to Solution 2, but since we are looking for the maximum window, we don’t shrink the window.

If the window is valid, we only increment the right edge.

Otherwise, we increment the left edge as well to shift the window as a whole.

// OJ: https://leetcode.com/problems/longest-continuous-subarray-with-absolute-diff-less-than-or-equal-to-limit/
// Time: O(N)
// Space: O(N)
// Ref: https://leetcode.com/problems/longest-continuous-subarray-with-absolute-diff-less-than-or-equal-to-limit/discuss/609771/JavaC%2B%2BPython-Deques-O(N)
class Solution {
public:
    int longestSubarray(vector<int>& A, int k) {
        deque<int> mx, mn;
        int i = 0, j = 0;
        for (; j < A.size(); ++j) {
            while (mx.size() && mx.back() < A[j]) mx.pop_back();
            while (mn.size() && mn.back() > A[j]) mn.pop_back();
            mx.push_back(A[j]);
            mn.push_back(A[j]);
            if (mx.front() - mn.front() > k) {
                if (A[i] == mx.front()) mx.pop_front();
                if (A[i] == mn.front()) mn.pop_front();
                ++i;
            }
        }
        return j - i;
    }
};

class Solution {
    public int longestSubarray(int[] nums, int limit) {
        if (nums == null || nums.length == 0)
            return 0;
        int maxLength = 0;
        int start = 0;
        TreeMap<Integer, Integer> map = new TreeMap<Integer, Integer>();
        int length = nums.length;
        for (int i = 0; i < length; i++) {
            int num = nums[i];
            int count = map.getOrDefault(num, 0) + 1;
            map.put(num, count);
            while (map.lastKey() - map.firstKey() > limit) {
                int prevNum = nums[start];
                int prevCount = map.get(prevNum) - 1;
                if (prevCount == 0)
                    map.remove(prevNum);
                else
                    map.put(prevNum, prevCount);
                start++;
            }
            maxLength = Math.max(maxLength, i - start + 1);
        }
        return maxLength;
    }
}

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

class Solution {
    public int longestSubarray(int[] nums, int limit) {
        TreeMap<Integer, Integer> tm = new TreeMap<>();
        int ans = 0, j = 0;
        for (int i = 0; i < nums.length; ++i) {
            tm.put(nums[i], tm.getOrDefault(nums[i], 0) + 1);
            while (tm.lastKey() - tm.firstKey() > limit) {
                tm.put(nums[j], tm.get(nums[j]) - 1);
                if (tm.get(nums[j]) == 0) {
                    tm.remove(nums[j]);
                }
                ++j;
            }
            ans = Math.max(ans, i - j + 1);
        }
        return ans;
    }
}

// OJ: https://leetcode.com/problems/longest-continuous-subarray-with-absolute-diff-less-than-or-equal-to-limit/
// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
    int longestSubarray(vector<int>& A, int k) {
        multiset<int> s{ A[0] };
        int i = 0, j = 1, N = A.size(), ans = 1;
        while (j < N) {
            s.insert(A[j++]);
            while (*s.rbegin() - *s.begin() > k) s.erase(s.find(A[i++]));
            ans = max(ans, j - i);
        }
        return ans;
    }
};

from sortedcontainers import SortedList


class Solution:
    def longestSubarray(self, nums: List[int], limit: int) -> int:
        sl = SortedList()
        ans = j = 0
        for i, v in enumerate(nums):
            sl.add(v)
            while sl[-1] - sl[0] > limit:
                sl.remove(nums[j])
                j += 1
            ans = max(ans, i - j + 1)
        return ans

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

class Solution(object):
    def longestSubarray(self, nums, limit):
        """
        :type nums: List[int]
        :type limit: int
        :rtype: int
        """
        N = len(nums)
        left, right = 0, 0
        res = 0
        visited = set()
        visited.add(nums[0])
        while right < N:
            if (len(visited) != 1):
                max_ = max(nums[left : right + 1])
                min_ = min(nums[left : right + 1])
                cur = max_ - min_
            else:
                cur = 0
            if cur > limit:
                left += 1
            else:
                visited.add(nums[right])
                res = max(res, right - left + 1)
                right += 1
        return res

func longestSubarray(nums []int, limit int) (ans int) {
	tm := treemap.NewWithIntComparator()
	j := 0
	for i, v := range nums {
		if x, ok := tm.Get(v); ok {
			tm.Put(v, x.(int)+1)
		} else {
			tm.Put(v, 1)
		}
		for {
			a, _ := tm.Min()
			b, _ := tm.Max()
			if b.(int)-a.(int) > limit {
				if x, _ := tm.Get(nums[j]); x.(int) == 1 {
					tm.Remove(nums[j])
				} else {
					tm.Put(nums[j], x.(int)-1)
				}
				j++
			} else {
				break
			}
		}
		ans = max(ans, i-j+1)
	}
	return
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

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;

    /**
   *
   * @param compareFn A compare function which returns boolean or number
   * @param leftBound defalut value is `-Infinity`
   * @param rightBound defalut value is `Infinity`
   * @description
   * create a `MultiSet`, compare elements using `compareFn`, which is increasing order by default.
   * @example
   * ```ts
   * interface Person {
        name: string
        age: number
    }

    const leftBound = {
        name: 'Alice',
        age: -Infinity,
    }

    const rightBound = {
        name: 'Bob',
        age: Infinity,
    }

    const sortedList = new TreapMultiSet<Person>(
        (a: Person, b: Person) => a.age - b.age,
        leftBound,
        rightBound
    )
   * ```
   */
    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);
    }
}

1438 - Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit

1438. Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit (Medium)

Solution 1. Sliding Window

Solution 2. Monotonous Deque

Solution 3. Monotonous Deque

All Problems

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1438. Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit (Medium)

Solution 1. Sliding Window

Solution 2. Monotonous Deque

Solution 3. Monotonous Deque

All Problems

All Solutions