<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) {
        ios_base::sync_with_stdio(false);
    cin.tie(NULL);
        deque<int> maxi, mini;
        int j = 0, i = 0;
        int ans = 0;
        
        for (i = 0; i < nums.size(); i++) {
            // Maintain the deque for the maximums
            while (!maxi.empty() && nums[i] > maxi.back()) {
                maxi.pop_back();
            }
            // Maintain the deque for the minimums
            while (!mini.empty() && nums[i] < mini.back()) {
                mini.pop_back();
            }
            // Add the current element to both deques
            maxi.push_back(nums[i]);
            mini.push_back(nums[i]);
            
            // Check if the current window is valid
            while (maxi.front() - mini.front() > limit) {
                if (maxi.front() == nums[j]) {
                    maxi.pop_front();
                }
                if (mini.front() == nums[j]) {
                    mini.pop_front();
                }
                j++;
            }
            
            // Update the answer with the length of the current valid window
            ans = max(ans, i - j + 1);
        }
        
        return ans;
    }
};
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