<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>

vector<int> Merge(vector<int> A, vector<int> B)
{
	vector<int> C;
	int i = 0, j = 0;
	while (true)
	{
		if (A[i] < B[j])
		{
			C.push_back(A[i]);
			i++;
			if (i == A.size()) break;
		}
		else
		{
			C.push_back(B[j]);
			j++;
			if (j == B.size()) break;
		}
	}
	while (i < A.size()) { C.push_back(A[i]); i++; }
	while (j < B.size()) { C.push_back(B[j]); j++; }
	return C;
}

void MergeSort(vector<int> * A)
{
	if ((*A).size() == 1) return;
	vector<int> B, C;
	for (int i = 0; i < (*A).size() / 2; i++) B.push_back((*A)[i]);
	for (int i = (*A).size() / 2; i < (*A).size(); i++) C.push_back((*A)[i]);
	MergeSort(&B);
	MergeSort(&C);
	(*A) = Merge(B, C);
	return;
}

class Solution {
public:
    int smallestDistancePair(vector<int>& nums, int k) {
        vector<int> distances;
        for (int i = 0; i <= 1000000; i++) distances.push_back(0);
        for (int i = 0; i < nums.size() - 1; i++) for (int j = i + 1; j < nums.size(); j++) distances[abs(nums[i] - nums[j])]++;
        int counter = 0;
        for (int i = 0; i < 1000000; i++) 
        {
            counter += distances[i];
            if (counter >= k) return i;
        } 
        return 0;
    }
};
169185