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

 

Solutions

Solution: Queue

• Time complexity: O(nlogn)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {number[]} nums
 * @param {number} limit
 * @return {number}
 */
const longestSubarray = function (nums, limit) {
  const queue = [];
  const size = nums.length;
  let max = (min = nums[0]);

  for (let index = 0; index < size; index++) {
    const value = nums[index];

    max = Math.max(value, max);
    min = Math.min(value, min);
    queue.push(value);
    if (max - min <= limit) continue;
    const leftValue = queue.shift();

    leftValue === max && (max = Math.max(...queue));
    leftValue === min && (min = Math.min(...queue));
  }
  return queue.length;
};