<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 right(vector<int> v,int i,int k) {
        int l = i, r = v.size() - 1;
        int res = -1;
        while(l <= r) {
            int mid = (l + r) / 2;
            if(v[mid] <= k) {
                res = mid;
                l = mid + 1;
            }
            else if(v[mid] > k) {
                r = mid - 1;
            }
        }
        return res;
    }

    int check(vector<int> v,int k) {
        int cnt = 0;
        for(int i=0;i<v.size() -1;i++) {
            // int l = left(v,i+1,v[i] + k);
            int r = right(v,i+1,v[i] + k);
            // cout << k << " " << i << " " << r << endl;
            if(r != -1) cnt += r - i ;
        }
        // cout << "cnt: " << cnt << endl;
        return cnt;
    }

    int smallestDistancePair(vector<int>& n, int k) {
        sort(n.begin(),n.end());
        int l  = 0, r = 1e6 + 1, rs = 0;
        while(l <= r) {
            int mid = (l + r) / 2;
            int res = check(n,mid);
            if(res >= k) {
                rs = mid;
                r = mid - 1;
            }
            else {
                l =  mid + 1;
            }
        }
        return rs;
    }
};
400238