<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) {
        // Can do a two-ptr approach w/ linear time solution
        // Keep track of the length between start/end ptrs, take max as solution.
        // Also keep track of min/max in current run. Key obs: 
        // When adding element, dont need to recompute if keeping track of max/min of current run
        if (nums.size() == 0) return 0;
        if (nums.size() == 1) return 1;
        int start = 0;
        int end = 1; // inclusive
        int res_max = 0;
        multiset<int> cur_min_max_mset;
        cur_min_max_mset.insert(nums[0]);
        while (start < (nums.size() - res_max) && end < nums.size()) {
            const int cur_min = *cur_min_max_mset.begin();
            const int cur_max = *cur_min_max_mset.rbegin();
            if (abs(nums[end]-cur_max) <= limit && abs(nums[end]-cur_min) <= limit) {
                // add this element to run and continue
                cur_min_max_mset.insert(nums[end]);
                end++;
                res_max = max(res_max, end - start);
                // cur_min_max_mset.insert(nums[end]);
            }
            else {
                cur_min_max_mset.erase(cur_min_max_mset.find(nums[start]));
                start++;
            }
            if (start >= end) {
                end = start + 1;
                if (end >= nums.size()) return res_max;
                cur_min_max_mset.insert(nums[start]);
            }
        }
        return res_max;
    }
};
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