<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 n = nums.size();

        int i = 0, j = 0, cur_max = nums[0], cur_min = nums[0];
        std::map<int, int> nums_count;
        nums_count[nums[0]] = 1;
        int ans = 0;

        while (i < n)
        {
            if (j < i)
            {
                j = i;
                nums_count.clear();
                nums_count[nums[i]] = 1;
            }
            while (j + 1 < n && 
            std::max(nums_count.rbegin()->first, nums[j + 1]) - 
            std::min(nums_count.begin()->first, nums[j + 1]) <= limit)
            {
                j++;
                if (nums_count.find(nums[j]) != nums_count.end())
                    nums_count[nums[j]] += 1;
                else
                    nums_count[nums[j]] = 1;
            }
            ans = std::max(ans, j - i + 1);
            if (nums_count[nums[i]] - 1 == 0)
                nums_count.erase(nums[i]);
            else
                nums_count[nums[i]] -= 1;
            i++;
        }

        return ans;
    }
};
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