<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 count(map<int,int>& m, vector<int>& nums, int x,int max){
        int cnt = 0;
        for(auto e : nums){
            int k = e+x;
            k = min(max,k);
            cnt+=m[k]-m[e];
        }
        return cnt;
    }
    int smallestDistancePair(vector<int>& nums, int k) {
        int n = nums.size();
        map<int,int> m;
        sort(nums.begin(),nums.end());
        int min = nums[0],max=nums[n-1];
        int i = 0;
        while(i<min){
            m[i]=0;
            i++;
        }
        m[i]=1;
        for(int j=1;j<n;j++){
            while(i<nums[j]){
                m[i]=m[nums[j-1]];
                i++;
            }
            m[i]=m[nums[j-1]]+1;
        }
        int cnt=0;
        for(int i=0;i<=max;i++){
            int x = m[i]-m[i-1];
            if(x>=2) cnt+=(x*(x-1))/2;
            // cout<<i<<" "<<m[i]<<endl;
        }
        int l=0,r=max-min,mid=0;
        while(l<=r){
            mid = (l+r)/2;
            int p = cnt+count(m,nums,mid,max);
            cout<<mid<<" "<<p<<endl;
            // if(p==k) return mid;
            if(p<k) l=mid+1;
            else r = mid-1;
        }
        return l;
    }
};
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