LeetCode #1438 — MEDIUM

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

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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 <= 10⁵
1 <= nums[i] <= 10⁹
0 <= limit <= 10⁹

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Sliding Window · Segment Tree · Monotonic Queue

Example 1

[8,2,4,7]
4

Example 2

[10,1,2,4,7,2]
5

Example 3

[4,2,2,2,4,4,2,2]
0

Core Insight

What unlocks the optimal approach

Use a sliding window approach keeping the maximum and minimum value using a data structure like a multiset from STL in C++.
More specifically, use the two pointer technique, moving the right pointer as far as possible to the right until the subarray is not valid (maxValue - minValue > limit), then moving the left pointer until the subarray is valid again (maxValue - minValue <= limit). Keep repeating this process.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1438: Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit
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;
    }
}

// Accepted solution for LeetCode #1438: Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit
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
}

# Accepted solution for LeetCode #1438: Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit
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

// Accepted solution for LeetCode #1438: Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit
struct Solution;
use std::collections::VecDeque;

impl Solution {
    fn longest_subarray(nums: Vec<i32>, limit: i32) -> i32 {
        let mut max_queue = VecDeque::new();
        let mut min_queue = VecDeque::new();
        let n = nums.len();
        let mut res = 0;
        let mut l = 0;
        for r in 0..n {
            while let Some(&last) = max_queue.back() {
                if last < nums[r] {
                    max_queue.pop_back();
                } else {
                    break;
                }
            }
            while let Some(&last) = min_queue.back() {
                if last > nums[r] {
                    min_queue.pop_back();
                } else {
                    break;
                }
            }
            max_queue.push_back(nums[r]);
            min_queue.push_back(nums[r]);
            if max_queue.front().unwrap() - min_queue.front().unwrap() > limit {
                if *max_queue.front().unwrap() == nums[l] {
                    max_queue.pop_front();
                }
                if *min_queue.front().unwrap() == nums[l] {
                    min_queue.pop_front();
                }
                l += 1;
            }
            res = res.max(r - l + 1);
        }
        res as i32
    }
}

#[test]
fn test() {
    let nums = vec![8, 2, 4, 7];
    let limit = 4;
    let res = 2;
    assert_eq!(Solution::longest_subarray(nums, limit), res);
}

// Accepted solution for LeetCode #1438: Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit
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);
    }
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n × k) time

O(1) space

For each starting index, scan the next k elements to compute the window aggregate. There are n−k+1 starting positions, each requiring O(k) work, giving O(n × k) total. No extra space since we recompute from scratch each time.

SLIDING WINDOW

O(n) time

O(k) space

The window expands and contracts as we scan left to right. Each element enters the window at most once and leaves at most once, giving 2n total operations = O(n). Space depends on what we track inside the window (a hash map of at most k distinct elements, or O(1) for a fixed-size window).

Shortcut: Each element enters and exits the window once → O(n) amortized, regardless of window size.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.