<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>

#define ll long long
#define all(v) v.begin(),v.end();
class Solution {
public:
    ll chk(ll p,map<ll,ll> &m,vector<ll> &pre)
    {
        ll c=0;
        for(auto it:m)
        {
            auto it2 = m.lower_bound(it.first+p);
            if(it2==m.end() || it2->first > it.first+p)
                it2--;
            c+=((it.second)*(pre[it2->first]-pre[it.first]))+((it.second*(it.second-1))/2);
        }
        return c;
    }
    int smallestDistancePair(vector<int>& v, int k) {
        
        map<ll,ll> m;
        vector<ll> pre(1e6+1,0);
        ll n = v.size();
        // sort(all(v));
        for(ll i=0;i<n;i++)
        {
             m[v[i]]++;
             pre[v[i]]++;
        }
        for(ll i=1;i<1e6+1;i++)
            pre[i]=pre[i]+pre[i-1];
        
        ll low=0,high=1e6;
        while(high-low > 1)
        {
            ll mid = (low+high)/2;
            ll got = chk(mid,m,pre);
            cout<<low<<" "<<high<<" "<<mid<<" "<<got<<endl;
            if(got >= k)
                high=mid;
            else
                low=mid;
        }
        if(chk(low,m,pre) >= k)
             return low;
        else
             return high;
       
            
    }
};
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