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:
    def longestSubarray(self, nums: List[int], limit: int) -> int:
        # HEAP SLIDING WINDOW SOLUTION
        max_heap, min_heap = [], []
        res = 0
        l = 0
        
        for r, a in enumerate(nums):
            heapq.heappush(max_heap, [-a, r])
            heapq.heappush(min_heap, [a, r])
            
            while -max_heap[0][0] - min_heap[0][0] > limit:
                # we need to kick all the elments before left pointer
                while max_heap[0][1] <= l:
                    heapq.heappop(max_heap)
                
                while min_heap[0][1] <= l:
                    heapq.heappop(min_heap)
                
                l += 1
                
            res = max(res, r - l + 1)
            
        return res

class Solution:
    def longestSubarray(self, nums: List[int], limit: int) -> int:
        # HEAP SLIDING WINDOW SOLUTION
        max_heap, min_heap = [], []
        res = 0
        l = 0
        
        for r, a in enumerate(nums):
            heapq.heappush(max_heap, [-a, r])
            heapq.heappush(min_heap, [a, r])
            
            while -max_heap[0][0] - min_heap[0][0] > limit:
                # we need to kick all the elments before left pointer
                while max_heap[0][1] <= l:
                    heapq.heappop(max_heap)
                
                while min_heap[0][1] <= l:
                    heapq.heappop(min_heap)
                
                l += 1
                
            res = max(res, r - l + 1)
            
        return res
