<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:
    void update(vector<int>& BIT,const int& num){
        for(int i=num;i<BIT.size();i+=(i&(-i)))BIT[i]++;
    }
    int query(vector<int>& BIT,const int& num){
        int ans=0;
        if(num<=0)return 0;
        for(int i=num;i>0;i-=(i&(-i)))ans+=BIT[i];
        return ans;
    }
    int c(const vector<int>& v,const int& diff){
        int ans=0;
        vector<int> BIT(1000002,0);
        int c=0;
        for(const auto&x:v){
            int tmp=query(BIT,x-diff);
            ans+=(c-tmp);
            update(BIT,x+1);
            c++;
        }
        return ans;
    }
    int smallestDistancePair(vector<int>& nums, int k){
        sort(nums.begin(),nums.end());
        int l=INT_MAX;
        for(int i=1;i<nums.size();i++)l=min(l,nums[i]-nums[i-1]);
        int r=nums[nums.size()-1]-nums[0];
        int ans=INT_MAX;
        while(l<=r){
            int m=l+(r-l)/2;
            int count = c(nums,m);
            //cout <<"$" << l << " " << r << " " << count<<endl;
            if(count>=k){
                ans=min(ans,m);
                r=m-1;
            }else{
                l=m+1;
            }
        }
        return ans;
    }
};
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