<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) {
        int i = 0, j = 0;
        int max = INT_MIN;
        int numMin = nums[0];
        int numMax = nums[0];
        int iMin = 0;
        int iMax = 0;
        priority_queue<pair<int, int>, vector<pair<int,int>>, greater<pair<int,int>>> min_q;
        priority_queue<pair<int, int>> max_q;

        while (j < nums.size()) {
            if (nums[j] > numMax) {
                numMax = nums[j];
                iMax = j;
            }
            if (nums[j] < numMin) {
                numMin = nums[j];
            }
            min_q.push(make_pair(nums[j], j));
            max_q.push(make_pair(nums[j], j));
            
            if (abs(numMin - numMax) <= limit) {
                if (j - i + 1 > max) {
                    max = j - i + 1;
                    if (max == nums.size()) {
                        return max;
                    }
                }
            }
            else {
                ++i; // window shortened
                if (iMin < i) {
                    while (!min_q.empty()) {
                        if (min_q.top().second >= i) {
                            numMin = min_q.top().first;
                            iMin = min_q.top().second;
                            break;
                        }
                        min_q.pop();
                    }
                }
                if (iMax < i) {
                    while (!max_q.empty()) {
                        if (max_q.top().second >= i) {
                            numMax = max_q.top().first;
                            iMax = max_q.top().second;
                            break;
                        }
                        max_q.pop();
                    }
                }
            }
            ++j; // window got bigger
        }
        
        return max;
    }
};
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