<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 smallestDistancePair(vector<int>& nums, int k) {
        
        vector<int>freq(1e6+1,0);
        vector<int>freqSum(1e6+1,0);
        int maxi=INT_MIN;

        int n=nums.size();

        for(int i=0;i<n;i++){
            for(int j=i+1;j<n;j++){
                int d=abs(nums[j]-nums[i]);
                maxi=max(maxi,d);
                freq[d]++;
            }
        }
        freqSum[0]=freq[0];
        for(int i=1;i<=maxi;i++){
            freqSum[i]+=freq[i];
            freqSum[i]+=freqSum[i-1];
        }
        int low=0,high=maxi;

        while(low<=high){
            int mid=low+(high-low)/2;
            // cout<<mid<<endl;
            if(freqSum[mid]<k){
                low=mid+1;
            }
            else if(freqSum[mid]-freq[mid]>=k){
                high=mid-1;
            }
            else if(freq[mid]==0){
                high=mid-1;
            }
            else{
                // cout<<freq[mid]<<" "<<freqSum[mid];
                return mid;
            }
        }
        return low;
    }
};
152681