<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:
        min_heap, max_heap = [], []
        max_diff = 0
        l = 0
        for i in range(len(nums)):
            # cur_len = i+1
            while min_heap and abs(nums[i] - min_heap[0][0]) > limit:
                _, j = heapq.heappop(min_heap)
                l = max(l, j+1)
                # cur_len = min(cur_len, i-j)
            
            while max_heap and abs(nums[i] + max_heap[0][0]) > limit:
                _, j = heapq.heappop(max_heap)
                l = max(l, j+1)
                # cur_len = min(cur_len, i-j)
            # print(i, l)
            max_diff= max(max_diff, i - l+1)
            heapq.heappush(min_heap, (nums[i], i))
            heapq.heappush(max_heap, (-nums[i], i))

        return max_diff
        
        
        # rolling_sum = 0
        # max_length = min(1, len(nums))
        # l, r = 0, 1

        # while r < len(nums):
        #     rolling_sum += nums[r] - nums[r-1]
        #     while l <= r and  abs(rolling_sum) > limit:
        #         max_length = max(max_length, r- l)
        #         rolling_sum -= (nums[l+1] - nums[l])
        #         l += 1
        #     max_length = max(max_length, r- l+1)
        #     r += 1
        # return max_length


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