<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

class Solution:
    def longestSubarray(self, nums: List[int], limit: int) -> int:
        l, r = 0, 0
        nums_len = len(nums)
        
        max_size = -1
        window = {}
        min_heap, max_heap = [], []
        while l < nums_len:
            while r < nums_len:
                entering = nums[r]
                r += 1

                window[entering] = window.get(entering, 0) + 1
                heapq.heappush(min_heap, entering)
                heapq.heappush(max_heap, -entering)

                # print(window)
                min_val = min_heap[0]
                while window.get(min_val, 0) <= 0:
                    heapq.heappop(min_heap)
                    min_val = min_heap[0]
                
                max_val = -max_heap[0]
                while window.get(max_val, 0) <= 0:
                    heapq.heappop(max_heap)
                    max_val = -max_heap[0]
                
                delta = max_val - min_val
                # print(nums[l:r], delta, max_val, min_val)
                if delta > limit: break

                cur_size = r-l
                max_size = max(max_size, cur_size)
            
            leaving = nums[l]
            window[leaving] -= 1
            l += 1
        
        return max_size


31760