Find K-th Smallest Pair Distance
<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>
Array; Two Pointers; Binary Search; Sorting
class Solution {
public:
    int smallestDistancePair(vector<int>& nums, int k) {
        sort(nums.begin(), nums.end());
        int left = 0, right = nums.back() - nums.front();
        while (left < right) {
            int mid = (left + right) >> 1;
            if (count(mid, k, nums) >= k)
                right = mid;
            else
                left = mid + 1;
        }
        return left;
    }

    int count(int dist, int k, vector<int>& nums) {
        int cnt = 0;
        for (int i = 0; i < nums.size(); ++i) {
            int target = nums[i] - dist;
            int j = lower_bound(nums.begin(), nums.end(), target) - nums.begin();
            cnt += i - j;
        }
        return cnt;
    }
};