<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) {
        deque<int> minQ;
        deque<int> maxQ;

        int begin = 0;
        int end = 0;
        int maxCount = 0;
        while (begin < nums.size())
        {
            while (end < nums.size())
            {
                if ((!minQ.empty() && abs(nums[end] - minQ.front()) > limit)
                    || (!maxQ.empty() && abs(nums[end] - maxQ.front()) > limit))
                {
                    break;
                }

                while (minQ.size() > 0 && nums[end] < minQ.back())
                {
                    minQ.pop_back();
                }
                minQ.push_back(nums[end]);

                while (maxQ.size() > 0 && nums[end] > maxQ.back())
                {
                    maxQ.pop_back();
                }
                maxQ.push_back(nums[end]);

                end++;
            }

            if (maxCount < end - begin)
                maxCount = end - begin;

            if (nums[begin] == minQ.front())
                minQ.pop_front();

            if (nums[begin] == maxQ.front())
                maxQ.pop_front();

            begin++;
        }

        return maxCount;
    }
};
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