<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 countpairs(int maxidistance,vector<int>&nums,vector<int>prefixcount,unordered_map<int,int>mp){
        int cnt=0;
        int n=nums.size();
        int i=0;
        while(i<n){
            int currval=nums[i];
            int currvalcnt=mp.at(currval);
            int largerpairs=prefixcount[currval+maxidistance]-prefixcount[currval];
            int samevaluepairs=currvalcnt*(currvalcnt-1)/2;
            cnt+=largerpairs*currvalcnt+samevaluepairs;
            while(i+1<n && nums[i]==nums[i+1]){
                i++;
            }
            i++;
        }
        return cnt;
    }
    int smallestDistancePair(vector<int>& nums, int k) {
        sort(nums.begin(),nums.end());
        int n=nums.size();
        int maxidistance=nums[n-1]*2;
        vector<int>prefixcount(maxidistance,0);
        unordered_map<int,int>mp;
        int j=0;
        for(int i=0;i<maxidistance;i++){
            while(j<n && nums[j]<=i){
                j++;
            }
            prefixcount[i]=j;
        }
        for(int i=0;i<n;i++){
            mp[nums[i]]++;
        }
        int low=0,high=nums[n-1];
        while(low<high){
            int mid=(low+high)/2;
            int cnt=countpairs(mid,nums,prefixcount,mp);
            if(cnt<k)low=mid+1;
            else high=mid;
        }
        return low;
    }
};
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