<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:
        
        # need to ensure difference between any two elements <= limit
        # maintain sliding window over arr
        # if we remove a maximum element from our window 
        # -> how can we update it with next largest element ??
        # 8, 5, 6, 2 -> limit = 4, how to update "largest" to 6 after removing 8
        # 7, 8, 5, 2 -> limit = 5, how to udpdate "largest
        # two heaps
        # iterate through nums, for every num add (num, idx) to min and max heap
        
        # two deques, one monotonically decreasing and the other monotonically increasing
        # add both nums to increasing queue and decreasing queue
        # largest element is at the front of decreasing 
        # smallest element is at the front of increasing
        # if decreasing[0] - increasing[0] > limit
        # remove elements from queue with the smallest idx at the front, update left pointer to that smallest index + 1
        decreasing_q, increasing_q = deque(), deque()
        left = 0
        res = 0

        for idx, num in enumerate(nums):
            while (len(decreasing_q) > 0 and decreasing_q[-1][0] < num):
                decreasing_q.pop()
            
            while (len(increasing_q) > 0 and increasing_q[-1][0] > num):
                increasing_q.pop()
            
            decreasing_q.append((num, idx))
            increasing_q.append((num, idx))

            while (decreasing_q[0][0] - increasing_q[0][0] > limit):
                if (decreasing_q[0][1] < increasing_q[0][1]):
                    left = decreasing_q[0][1] + 1
                    while (decreasing_q[0][1] < left):
                        decreasing_q.popleft()
                else:
                    left = increasing_q[0][1] + 1
                    while (increasing_q[0][1] < left):
                        increasing_q.popleft()
            
            res = max(res, idx - left + 1)
        return res












        
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