<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:
        # 1) brute force, find every subarray and compute max - min
        # stack = [0]
        # for i in range(0, len(nums), 1):
        #     for j in range(i, len(nums), 1):
        #         subArr = nums[i:j+1]
        #         if max(subArr) - min(subArr) <= limit:
        #             if len(subArr) > stack[-1]:
        #                 stack[-1] = len(subArr)

        # return stack[0]

        # 2) optimal solution using 2 monotonic queues one for max and one for min
        minQueue = deque([]) # (idx, val) | guarantees that minQueue[0] is min in currWindow
        maxQueue = deque([]) # (idx, val) | gurantees that maxQueue[0] is max in currWindow

        l = 0
        maxSeq = 0
        for r,v in enumerate(nums):
            # while currVal < minQueue, pop(), currVal append at the end
            while minQueue and nums[r] < minQueue[-1][1]:
                minQueue.pop()
            minQueue.append((r, nums[r]))
            # while currVal > maxQueue, pop(), currVal append at the end
            while maxQueue and nums[r] > maxQueue[-1][1]:
                maxQueue.pop()
            maxQueue.append((r, nums[r]))
        
            # check if left pointer needs shifting | currSum > limit
            if maxQueue[0][1] - minQueue[0][1] > limit:
                l += 1
                if minQueue[0][0] < l:
                    minQueue.popleft()
                if maxQueue[0][0] < l:
                    maxQueue.popleft()
            
            maxSeq = max(maxSeq, (r - l + 1))
        
        return maxSeq

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