<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>

from collections import deque
from sortedcontainers import SortedList

class Solution:
    def longestSubarray(self, nums: List[int], limit: int) -> int:

        # [l, r]; [0, ri) for ri = 1..n
        # 
        # [l + 1, ..
        # [5, 1, 1, 4, 1, 1, 3, 2, 1]

        # limit = 5
        # [10,1,2,4,7,2]
        # [10] ans = 1, mi = 10, ma = 10
        # -- [10, 1]; mi = 10 -> 1, ma = 10
        # [1], ans = 1
        # [1, 2], ans = 2
        # [1, 2, 4], ans = 3; min = 1, max = 4
        # [2, 4, 7], ans = 3; min = 2, max = 7
        # [2, 4, 7, 2], ans = 4; min = 2; max = 7

        # limit = 0
        # nums = [4,2,2,2,4,4,2,2]

        # [4], ans = 1
        # [2], ans = 1
        # [2, 2], ans = 2,
        # ...
        # [4], ans = 4

        # [ )

        # l 
        # nums = [8, 1, 1, 1, 1, 9, 4]
        l = 0
        r = 1
        n = len(nums)
        ans = 0

        cur = SortedList([])
        while r <= n:
            cur.add(nums[r - 1])
            mi = next(iter(cur))
            ma = next(reversed(cur))

            if ma - mi <= limit:
                ans = max(ans, r - l)
            else:
                while ma - mi > limit:
                    cur.remove(nums[l])
                    mi = next(iter(cur))
                    ma = next(reversed(cur))
                    l += 1

            r += 1

        return ans



        
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