<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) {
        unordered_map<int, int> count;
        for (int n : nums) count[n]++;
        vector<int> set;
        for (auto it = count.begin(); it != count.end(); it++) {
            set.push_back(it->first);
            k -= (it->second * (it->second - 1)) / 2;
        }
        if (k <= 0) return 0;
        // sort(set.begin(), set.end());
        int n = set.size();
        vector<int> distances(1000000);
        for (int i = 0; i < n - 1; i++) {
            for (int j = i + 1; j < n; j++) {
                int a = set[i];
                int b = set[j];
                int d = abs(a - b);
                int c = count[a] * count[b];
                distances[d] += c;
            }
        }
        for (int i = 0; i < 1000000; i++) {
            k -= distances[i];
            if (k <= 0) return i;
        }
        return 0;
    }
};

70163