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

        maxmum=[]
        hq.heapify(maxmum)

        minmum=[]
        hq.heapify(minmum)

        i=0
        j=0
        prevJ=-1

        longest=0
        s=set({})

        while(i<len(nums)):

            if(prevJ!=j):
                hq.heappush(maxmum,(-nums[j],j))
                hq.heappush(minmum,(nums[j],j))

            if(j+1<len(nums)):
                prevJ=j
                minn=minmum[0][0]
                maxx=abs(maxmum[0][0])
                
                if(nums[j+1]<minn):
                    minn=nums[j+1]
                if(nums[j+1]>maxx):
                    maxx=nums[j+1]
                
                if(abs(maxx-minn)<=limit):
                    j=j+1
                else:
                    if(j-i+1>longest):
                        longest=j-i+1

                    if(abs(maxmum[0][0])==nums[i]):
                        hq.heappop(maxmum)
                    if(minmum[0][0]==nums[i]):
                        hq.heappop(minmum)

                    s.add(i)     
                    if(i==j):
                        j=j+1
                        prevJ=-1
                    i=i+1

            else:
                prevJ=j
                if(j-i+1>longest):
                    longest=j-i+1

                if(abs(maxmum[0][0])==nums[i]):
                    hq.heappop(maxmum)
                if(minmum[0][0]==nums[i]):
                    hq.heappop(minmum)
                
                s.add(i)
                if(i==j):
                    j=j+1
                    prevJ=-1
                i=i+1
            
            while(len(maxmum)>0 and abs(maxmum[0][1]) in s):
                hq.heappop(maxmum)
            
            while(len(minmum)>0 and minmum[0][1] in s):
                hq.heappop(minmum)

        return longest        
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