<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) {
        int sv = INT_MAX;
        int lv = INT_MIN;
        for(int n: nums) {
            sv = min(sv, n);
            lv = max(lv, n);
        }
        int h = lv - sv;
        int l = 0;
        int cand = INT_MAX;
        while(l <= h) {
            int mid = l + (h-l)/2;
            int t = 0;
            vector<int> bit(1e6+2, 0);
            // unordered_set<int> ms;
            bool f = false;
            for(int n: nums) {
                t += countRange(bit, min(n+mid, (int)1e6)) - countRange(bit, max(n-mid-1, -1));
                // if(ms.count(n-mid) > 0 || ms.count(n+mid) > 0) f = true;
                if(t >= k) break;
                update(bit, n);
                // ms.insert(n);
            }
            if(t >= k) {
                cand = min(cand, mid);
                h = mid-1;
            } else {
                l = mid+1;
            }
        }
        return cand;
    }
    int countRange(vector<int> &bit, int n) {
        int ind = n + 1;
        int s = 0;
        while(ind > 0) {
            s += bit[ind];
            ind -= (ind & -ind);
        }
        return s;
    }
    void update(vector<int> &bit, int n) {
        int ind = n + 1;
        while(ind <= 1e6+1) {
            bit[ind] += 1;
            ind += (ind & -ind);
        }
        return;
    }
};
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