<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 {
   
public:
    //here I have used pair to handle duplicate values and can be categoried by index
    int longestSubarray(vector<int>& nums, int limit) {
        int i=0,j=0,n=nums.size(),mx,mn,res=0;
        priority_queue<pair<int,int>> mxhp;   //to find max value in window
        priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>> mnhp; 
        //to find min value in window
        while(i<n) //until we reach to end
        {
            while(j<n) 
            {
                mxhp.push({nums[j],j}); //push value and index in max heap
                mnhp.push({nums[j],j}); //push value and index in min heap
                mx=mxhp.top().first;    //store max value of window
                mn=mnhp.top().first;    //store min value of window
                if(mx-mn>limit)         //if we crosses limit difference , then now we have to change window     
                {   
                    while(!mxhp.empty()&&mxhp.top().second<=i) mxhp.pop();//remove ith or before element from heap
                    while(!mnhp.empty()&&mnhp.top().second<=i) mnhp.pop();//remove ith or before element from heap 
                    j++;break;          
                    //increment window size and break window, because we have stored jth 
                    //index value , next time we will store (j+1)th index value
                }
                res=max(j-i+1,res);     //update window length 
                j++;                    //increase window size
            } 
            i++;    //decrease windowsize  from starting
        } 
        return res; //return final maximum window size
    }
};
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