<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:
        
        return _use_deque(nums, limit)


def _use_deque(nums: List[int], limit: int):
    max_q, min_q = deque(), deque()

    left = 0

    for x in nums:
        while max_q and max_q[-1] < x:
            max_q.pop()
        max_q.append(x)

        while min_q and min_q[-1] > x:
            min_q.pop()
        min_q.append(x)

        if max_q[0] - min_q[0] > limit:
            if max_q[0] == nums[left]:
                max_q.popleft()

            if min_q[0] == nums[left]:
                min_q.popleft()
            
            left += 1

    return len(nums) - left

def _use_heap(nums: List[int], limit: int):

    max_heap, min_heap = [], []
    ans = 0
    left = 0
    for i, x in enumerate(nums):

        heapq.heappush(max_heap, [-x, i])
        heapq.heappush(min_heap, [x, i])

        while -max_heap[0][0] - min_heap[0][0] > limit:
            left = min(max_heap[0][1], min_heap[0][1]) + 1
            while max_heap[0][1] < left:
                heapq.heappop(max_heap)
            while min_heap[0][1] < left:
                heapq.heappop(min_heap)
        if (_len := i - left + 1) > ans:
            ans = _len
    return ans
        
272