<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>

class Solution:
    def longestSubarray(self, nums: List[int], limit: int) -> int:
        # max(a, i) min(b, j) => valid
        # [i, j] => valid

        # 10 1 3 5 7  lim = 2
        #      l   r
        # max: [7] 1 2 3
        # min: [5] 10 5 7
        # if max - min <= limit 
        # => r += 1
        # else: 
        # l += 1 and !isInWindow(minH, maxH) => pop()

        # 8 2 4 7 limit = 4
        #   lr
        # num = 2
        # max = [2] 
        # min = [2] 8

        # maxHeap = []
        # minHeap = []
        # l = 0
        # r = 0

        # # max khong nam trong window => pop max
        # for i, num in enumerate(nums):
        #     while maxHeap and minHeap and :
        #         if abs(maxHeap[0] - minHeap[0]) <= limit:
        #             r += 1
        #         else:
        #             l += 1
        #             if maxHeap[0][1] < l:
        #                 heapq.heappop(maxHeap)
        #             if abs(maxHeap[0] - minHeap[0]) > limit and l == minHeap[0][1]:
        #                 heapq.heappop(minHeap)
                    
        #     heapq.heappush(maxHeap, (num, i))
        #     heapq.heappush(minHeap, (num, i))
        # return -1
        minH = []
        maxH = []
        L = 0
        R = 0
        res = 0
        while R < len(nums):
            heapq.heappush(minH, (nums[R], R))
            heapq.heappush(maxH, (-nums[R], R))
            R += 1

            # if -maxH[0][0] - minH[0][0] <= limit:
            #     res = max(res, maxH[0][1] - minH[0][1])
            #     continue

            while -maxH[0][0] - minH[0][0] > limit: 
                L += 1

                while maxH and maxH[0][1] < L:
                    heapq.heappop(maxH)

                while minH and minH[0][1] < L:
                    heapq.heappop(minH)
            res = max(res, R - L)
        return res

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