<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) {
        sort(nums.begin(), nums.end());
        vector<int> dists; 
        dists.reserve(nums.size() * nums.size()-1);
        for (int i=0; i < nums.size(); ++i) {
            int ni = nums[i];
            for (int j=i+1; j < nums.size(); ++j) {
                int nj = nums[j];
                dists.push_back(nj - ni);
            }
        }
        // for (int d: dists) { cout << d << ", "; } cout << "\n";
        vector<int>::iterator itr = dists.begin() + k - 1;
        nth_element(dists.begin(), itr, dists.end());
        return *itr;
    }

#if 0
    int smallestDistancePair(vector<int>& nums, int k) {
        priority_queue<int> heap;
        for (int i=0; i < nums.size(); ++i) {
            for (int j=i+1; j < nums.size(); ++j) {
                int d = abs(nums[i] - nums[j]);
                if (heap.size() < k) {
                    heap.push(d);
                    continue;
                } else {
                    int largest = heap.top();
                    if (d < largest) {
                        heap.pop();
                        heap.push(d);
                    }
                }
            }
        }
        return heap.top();
    }
#endif
};
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