<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 solve(vector<int>& nums,vector<int> prefixcount,unordered_map<int,int> valuecount,int distance){
    int i,count;count=0;i=0;
while(i<nums.size()){
    int pairwithdistance=prefixcount[nums[i]+distance]-prefixcount[nums[i]];
    int valuecountdistance=valuecount[nums[i]];
    int diffzero=valuecountdistance*(valuecountdistance-1)/2;
    count+=(pairwithdistance*valuecountdistance)+diffzero;
    while(i+1<nums.size()&&nums[i]==nums[i+1]){
        i++;
    }
    i++;
}
return count;
}
    int smallestDistancePair(vector<int>& nums, int k) {
        sort(nums.begin(),nums.end());
        int maxelement=nums[nums.size()-1];
        vector<int> prefixcount(maxelement*2,0);
        unordered_map<int,int> valuecount;
        int index;index=0;
        for(int i=0;i<maxelement*2;i++){
              while(index<nums.size()&&nums[index]<=i){
                index++;
              }
              prefixcount[i]=index;
        }

        for(int i=0;i<nums.size();i++){
            valuecount[nums[i]]++;
        }
        int low,high;low=0;high=maxelement;
       
        while(low<high){
            int mid=low+(high-low)/2;
            int count=solve(nums,prefixcount,valuecount,mid);
            if(count<k){
                low=mid+1;
            }else{
                high=mid;
            }
        }
        return low;
    }
};
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