human
I found an interesting problem on https://leetcode.com/problems/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

I tried it in Python, but could not do it. Can you solve it?
assistant
#Here is the Python code you requested.

class Solution:
    def longestSubarray(self, nums: List[int], limit: int) -> int:
        maxque = collections.deque([])
        minque = collections.deque([])
        start = end = 0
        count = 0
        while end < len(nums):
            while len(maxque) and nums[end] > nums[maxque[-1]]: maxque.pop()
            while len(minque) and nums[end] < nums[minque[-1]]: minque.pop()
            maxque.append(end)
            minque.append(end)
            if nums[maxque[0]] - nums[minque[0]] > limit:
                if maxque[0] == start:
                    maxque.popleft()
                if minque[0] == start:
                    minque.popleft()
                start += 1
                end += 1
            else:
                end += 1

        return len(nums) - start