<p>The <strong>distance of a pair</strong> of integers <code>a</code> and <code>b</code> is defined as the absolute difference between <code>a</code> and <code>b</code>.</p>

<p>Given an integer array <code>nums</code> and an integer <code>k</code>, return <em>the</em> <code>k<sup>th</sup></code> <em>smallest <strong>distance among all the pairs</strong></em> <code>nums[i]</code> <em>and</em> <code>nums[j]</code> <em>where</em> <code>0 &lt;= i &lt; j &lt; nums.length</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,3,1], k = 1
<strong>Output:</strong> 0
<strong>Explanation:</strong> Here are all the pairs:
(1,3) -&gt; 2
(1,1) -&gt; 0
(3,1) -&gt; 2
Then the 1<sup>st</sup> smallest distance pair is (1,1), and its distance is 0.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,1,1], k = 2
<strong>Output:</strong> 0
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,6,1], k = 3
<strong>Output:</strong> 5
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == nums.length</code></li>
	<li><code>2 &lt;= n &lt;= 10<sup>4</sup></code></li>
	<li><code>0 &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>
	<li><code>1 &lt;= k &lt;= n * (n - 1) / 2</code></li>
</ul>

class Solution {
public:
    int smallestDistancePair(vector<int>& nums, int k)
    {
//k = nums.size() * nums.size();//KKKKKKKKKKKKKKKKKKKKKKKKKKKK
        std::priority_queue<int, std::vector<int>
        //, std::greater<int> 
        > pq;

        sort(nums.begin(), nums.end());
//////for (auto o : nums) {cout<<o<<" ";} cout<<endl;

        for (int d = 1; d < nums.size() - 1; ++d)
        {
            //bool cont = false;
            for (int i = 0; i+d < nums.size(); ++i)
            {
                int n = i+d;
                //cont = true;
                int o = abs(nums[i] - nums[n]);

                if (pq.size() < k)
                    pq.push(o);
                else
                {
                    auto pq_top = pq.top();
                    if (pq_top == o   &&   k < pq.size())
                    {
                        pq.push(o);
                    }
                    else
                    if (pq_top > o)
                    {
                        pq.push(o);
                        
                        if (pq.size() > k)
                            pq.pop();
                    }
                }
            }

            //if (!cont) break;
            //if (k == pq.size()   &&   0 == pq.top()) break;
        }

        int ret = pq.top();

//for (;pq.size();) {cout<<pq.top()<<" "; pq.pop();}
        
        return ret;
    }
};
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