<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:
        left, right = 0, 0
        max_window = []
        min_window = []
        max_removed = {} # value: count
        min_removed = {}

        out = 0
        while right < len(nums):
            heappush(max_window, -nums[right])
            heappush(min_window, nums[right])
            
            # print(nums[right])
            while abs(-max_window[0] - min_window[0]) > limit:
                value_to_remove = nums[left]
                max_removed[value_to_remove] = max_removed.get(value_to_remove, 0) + 1
                min_removed[value_to_remove] = min_removed.get(value_to_remove, 0) + 1

                while max_removed.get(-max_window[0], 0) > 0:
                    max_removed[-max_window[0]] -= 1
                    heappop(max_window)
                while min_removed.get(min_window[0], 0) > 0:
                    min_removed[min_window[0]] -= 1
                    heappop(min_window)
                    # min_removed[min_window[0]] -= 1
                
                left += 1
            
            out = max(out, right - left + 1)

            right += 1
        
        return out

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