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

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 <= 10^5

• 1 <= nums[i] <= 10^9

• 0 <= limit <= 10^9

class Solution {
    public int longestSubarray(int[] A, int limit) {
        Deque<Integer> maxDQ = new ArrayDeque<>();
        Deque<Integer> minDQ = new ArrayDeque<>();
        int start = 0, end, ans = 0;
        for (end = 0; end < A.length; ++end) {
            while (!maxDQ.isEmpty() && A[end] > maxDQ.peekLast())
                maxDQ.pollLast();
            while (!minDQ.isEmpty() && A[end] < minDQ.peekLast())
                minDQ.pollLast();
            maxDQ.add(A[end]);
            minDQ.add(A[end]);
            while (start < end && maxDQ.peek() - minDQ.peek() > limit) {
                if (maxDQ.peek() == A[start])
                    maxDQ.poll();
                if (minDQ.peek() == A[start])
                    minDQ.poll();
                start++;
            }
            ans = Math.max(ans, end - start + 1);
        }
        return ans;
    }
}