<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());

        int max_dist = nums.back() - nums.front();

        vector<int> cnt(nums.back() + 1 + max_dist);
        vector<int> occ(nums.back() + 1 + max_dist);
        for (int num : nums) {
            ++cnt[num];
            ++occ[num];
        }
        for (int i = 1; i < nums.back() + 1 + max_dist; ++i) {
            cnt[i] += cnt[i - 1];
        }

        int left = 0;
        int right = max_dist;
        while (left < right) {
            int mid = left + (right - left) / 2;
            int cnt_dist = helper(nums, cnt, occ, mid);
            if (cnt_dist < k) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        return left;
    }

private:
    int helper(vector<int>& nums, vector<int>& cnt, vector<int>& occ,
               int target) {
        int res = 0;
        int n = nums.size();
        for (int i = 0; i < n; ++i) {
            int x = nums[i];
            int y = target + nums[i];

            res += occ[x] * (cnt[y] - cnt[x]);
            res += occ[x] * (occ[x] - 1) / 2;

            while (i < n && nums[i] == x) {
                ++i;
            }
            --i;
        }
        return res;
    }
};
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