<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
        res = size = 0
        window = MonotonicQueue()
        while right < len(nums):
            window.push(nums[right])
            right += 1
            size += 1
            while window.maxq and window.minq and window.maxVal() - window.minVal() > limit:
                window.pop()
                left += 1
                size -= 1
            res = max(res, size)
        return res 
    
class MonotonicQueue:
    def __init__(self):
        self.maxq = deque()
        self.minq = deque()
        self.q = deque()
        
    def push(self, n):
        self.q.append(n)
        while self.maxq and self.maxq[-1] < n:
            self.maxq.pop()
        self.maxq.append(n)  
        while self.minq and self.minq[-1] > n:
            self.minq.pop()
        self.minq.append(n)
        
    def maxVal(self):
        if not self.maxq:
            return None
        return self.maxq[0]
    
    def minVal(self):
        if not self.maxq:
            return None
        return self.minq[0]
    
    def pop(self):
        if len(self.q) == 0:
            return None
        val = self.q.popleft()
        if self.maxq[0] == val:
            self.maxq.popleft()
        if self.minq[0] == val:
            self.minq.popleft()
        return val
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