<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 {
    int countLessMid(vector<int>& v, int mid) {
        int n = v.size();
        int cnt = 0;
        for(int i = 0; i < n;i++) {
            int until = v[i] + mid;
            int idx = upper_bound(v.begin(), v.end(), until) - v.begin();
            idx -= 1;
            if(idx > i)
                cnt += (idx - i);
        }
        return cnt;
    }

    int countEqualMid(vector<int>& v, unordered_map<int, int>& frq, int mid) {
        int n = v.size();
        int cnt = 0;
        if(mid == 0)
            return v.size();
        for(int i = 0; i < n;i++) {
            int until = v[i] + mid;
            cnt += frq[until];
        }
        return cnt;
    }
public:
    int smallestDistancePair(vector<int>& nums, int k) {
        int n = nums.size();
        unordered_map<int, int> frq;
        sort(nums.begin(), nums.end());
        for(auto &it: nums)
            frq[it] += 1;
        int left = 0, right = 1000000, mid = -1, res = -1;
        while(left <= right) {
            mid = left + (right - left) / 2;
            int less = countLessMid(nums, mid - 1);
            int equal = countEqualMid(nums, frq, mid);
            // cout << mid << ' ' << less << ' ' << equal << endl;
            if(less >= k)
                right = mid - 1;
            else if(less + equal < k)
                left = mid + 1;
            else {
                res = mid;
                break;
            }
        }
        return res;
    }
};
26153