<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 n;
    pair<int,int> steq(int dist,vector<int>& nums){
        map<int,int> freq;
        for(auto it: nums){
            freq[it]++;
        };
        pair<int,int> ans;
        for(int i = 0;i<n;i++){
            int it = nums[i];
            int idx = upper_bound(nums.begin(),nums.end(),it-dist)-nums.begin();
            int lt = i-idx;
            int eq = freq[it-dist];
            ans.first += lt;
            ans.second += eq;
        };
        ans.first++;
        ans.second += ans.first;
        ans.second--;
        return ans;
    };
    
    int smallestDistancePair(vector<int>& nums, int k) {
        n = nums.size();
        sort(nums.begin(),nums.end());
        int l = 0, r = (nums[n-1] -nums[0]);
        
        // cout<<"rs.first "<<rs.first<<endl;
        // cout<<"rs.second "<<rs.second<<endl;
    
        while(l<=r){
            int mid = (l+r)/2;
            pair<int,int> rs = steq(mid,nums);
            // cout<<"\n----------\nl and r is "<<l<<" "<<r<<endl;
            // cout<<"mid is "<<mid<<endl;
            // cout<<"range of mid is "<<rs.first<<" "<<rs.second<<endl;
            if(rs.first<=k && k<=rs.second){
                return mid;
            };
            if(rs.second<k){
                l = mid+1;
            }
            else{
                r = mid-1;
            };   
        };
        return l;
        
    }
};
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