<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:
   // Function to find the longest subarray with elements within a specific limit
int longestSubarray(vector<int>& nums, int limit) {
    // Initialize variables
    int i = 0, j = 0; // Pointers for sliding window (i - start, j - end)
    int n = nums.size(); // Size of the input vector
    int max_size = 0; // Stores the length of the longest subarray found so far

    // Initialize two priority queues
    // - maxh (max heap): Stores {element, index} pairs, ordered with element in descending order
    // - minh (min heap): Stores {element, index} pairs, ordered with element in ascending order
    priority_queue<pair<int, int>> maxh;
    priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> minh;

    // Sliding window loop
    while (j < n) {
        // Add the current element to both heaps
        maxh.push({nums[j], j}); // Add to max heap
        minh.push({nums[j], j}); // Add to min heap

        // Maintain the window by removing elements outside the current window
        while (maxh.top().second < i) {
            maxh.pop(); // Remove elements from max heap with index less than i
        }
        while (minh.top().second < i) {
            minh.pop(); // Remove elements from min heap with index less than i
        }

        // Check if the difference between maximum and minimum elements is within the limit
        if (maxh.top().first - minh.top().first <= limit) {
            // Update the maximum subarray size if the current window is larger
            max_size = max(max_size, j - i + 1);
            j++; // Expand the window by incrementing j
        } else {
            // Shrink the window from the beginning (increment i)
            i++;
        }
    }

    return max_size; // Return the length of the longest subarray found
}
};
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