<p>Given an array of integers <code>nums</code> and an integer <code>limit</code>, return the size of the longest <strong>non-empty</strong> subarray such that the absolute difference between any two elements of this subarray is less than or equal to <code>limit</code><em>.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [8,2,4,7], limit = 4
<strong>Output:</strong> 2 
<strong>Explanation:</strong> All subarrays are: 
[8] with maximum absolute diff |8-8| = 0 &lt;= 4.
[8,2] with maximum absolute diff |8-2| = 6 &gt; 4. 
[8,2,4] with maximum absolute diff |8-2| = 6 &gt; 4.
[8,2,4,7] with maximum absolute diff |8-2| = 6 &gt; 4.
[2] with maximum absolute diff |2-2| = 0 &lt;= 4.
[2,4] with maximum absolute diff |2-4| = 2 &lt;= 4.
[2,4,7] with maximum absolute diff |2-7| = 5 &gt; 4.
[4] with maximum absolute diff |4-4| = 0 &lt;= 4.
[4,7] with maximum absolute diff |4-7| = 3 &lt;= 4.
[7] with maximum absolute diff |7-7| = 0 &lt;= 4. 
Therefore, the size of the longest subarray is 2.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [10,1,2,4,7,2], limit = 5
<strong>Output:</strong> 4 
<strong>Explanation:</strong> The subarray [2,4,7,2] is the longest since the maximum absolute diff is |2-7| = 5 &lt;= 5.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [4,2,2,2,4,4,2,2], limit = 0
<strong>Output:</strong> 3
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>0 &lt;= limit &lt;= 10<sup>9</sup></code></li>
</ul>

import heapq
# import collections

class Solution:
    def longestSubarray(self, nums: List[int], limit: int) -> int:
        if not nums:
            return nums
        
        size = len(nums)

        # Heap to keep the min at each window
        heap_min = [(nums[0], 0)]
        heap_max = [(-nums[0], 0)]
        max_length = 1
        left = 0
        for right in range(1, size):
            while heap_min and abs(heap_min[0][0] - nums[right]) > limit:
                left = heap_min[0][1] + 1
                while heap_min and heap_min[0][1] < left: # Pop older content before the min value
                    heapq.heappop(heap_min)

            while heap_max and abs(-heap_max[0][0] - nums[right]) > limit:
                left = max(left, heap_max[0][1] + 1) # if heap_max is after the current left
                while heap_max and heap_max[0][1] < left:
                    heapq.heappop(heap_max)

            # Remove invalid min & max values outside of this window
            while heap_min and heap_min[0][1] < left:
                heapq.heappop(heap_min)
            while heap_max and heap_max[0][1] < left:
                heapq.heappop(heap_max)

            heapq.heappush(heap_min, (nums[right], right))
            heapq.heappush(heap_max, (-nums[right], right))
            max_length = max(max_length, right - left + 1)
        
        return max_length


# two deques        
        # global_max_len = 0
        # l = 0
        # r = 0
        # min_deque = collections.deque()
        # max_deque = collections.deque()

        # for r in range(len(nums)):
        #     while min_deque and min_deque[-1] > nums[r]:
        #         min_deque.pop()
        #     min_deque.append(nums[r])

        #     while max_deque and max_deque[-1] < nums[r]:
        #         max_deque.pop()
        #     max_deque.append(nums[r])

        #     while max_deque[0] - min_deque[0] > limit:
        #         if min_deque[0] == nums[l]:
        #             min_deque.popleft()
        #         if max_deque[0] == nums[l]:
        #             max_deque.popleft()
        #         l += 1
            
        #     global_max_len = max(r - l + 1, global_max_len)

        # return global_max_len

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