<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(map<int,int> mp,int target,int k)
    {
        int count = 0;
        int curr = 0;
        for(auto i = mp.begin(),j = mp.begin();i!=mp.end();++i)
        {
            while(j->first<i->first-target)
            {
                curr-=j->second;
                ++j;
            }
            count+=curr*i->second;
            count+=(i->second-1)*i->second/2;
            curr+=i->second;
            if(count>=k)
                return k;
        }
        return count;
    }
    int smallestDistancePair(vector<int>& nums, int k) {
        map<int,int> mp;
        for(auto n :nums)
            mp[n]++;
        int l = 0;
        int r = 10e6;
        int res = -1;
        while(l<=r)
        {
            int mid = (l+r)/2;
            if(countPairs(mp,mid,k)<k)
                l = mid+1;
            else
            {
                res = mid;
                r = mid-1;
            }         
        }
        return res;
    }
};
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