<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 int
class fenwick{
    vector<ll>sum;
    public:
    fenwick(ll n){
        sum.resize(n+1,0);
    }
    void add(ll idx,ll e){
        idx++;
        ll n=sum.size()-1;
        for(;idx<=n;idx+=(idx&(-idx))){
            sum[idx]+=e;
        }
    }
    //find sum from 0 upto idx 
    ll query(ll idx){
        if(idx<0) return 0; 
        ll ans=0;
        idx++;
        for(;idx>0;idx-=(idx&(-idx))) ans+=sum[idx];
        return ans;
    }
};

class Solution {

public:
    int smallestDistancePair(vector<int>& nums, int k) {
        sort(nums.begin(),nums.end());
        fenwick fw(1000001);
        for(auto t:nums) fw.add(t,1);
        auto possible=[&](int mid){
            int cnt=0;
            for(auto t:nums){
                fw.add(t,-1);
                int v=min(1000000,t+mid);
                cnt+=(fw.query(v)-fw.query(t-1)); 
            } 
            for(auto t:nums) fw.add(t,1);
            return cnt>=k;
        };
        int ans=0;
        int low=0,high=1e6;
        while(low<=high){
            int mid=(low+high)>>1;
            if(possible(mid)) ans=mid,high=mid-1;
            else low=mid+1;
        }
        return ans;
    }
};
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