<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:
    int longestSubarray(vector<int>& nums, int limit) {
        if(nums.size()==1) return 1;
        int s=0,e=1,res=1,n=nums.size();
        unordered_map<int,int> mp;
        priority_queue<int> mx;
        priority_queue<int,vector<int>,greater<int>> mi;
        mx.push(nums[0]);
        mx.push(nums[1]);
        mi.push(nums[0]);
        mi.push(nums[1]);
        mp[nums[0]]=0;
        mp[nums[1]]=1;
        if(abs(mx.top()-mi.top())<=limit) res=max(res,e-s+1);
        while(e<n){
            while(!mi.empty() && mp[mi.top()]<s) mi.pop();
            while(!mx.empty() && mp[mx.top()]<s) mx.pop();
            // cout<<s<<","<<e<<": "<<mx.top()<<","<<mi.top()<<endl; 
            if(abs(mx.top()-mi.top())>limit){
                if(mp[mx.top()]<mp[mi.top()]){
                    s=mp[mx.top()]+1;
                    mx.pop();
                }else{
                    s=mp[mi.top()]+1;
                    mi.pop();
                }
            }else{
                res=max(res,e-s+1);
                e++;
                if(e==n) return res;
                mi.push(nums[e]);
                mx.push(nums[e]);
                mp[nums[e]]=e;
            }
        }
        return res;
    }
};
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