<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) {
        sort(nums.begin(), nums.end());
        int t = nums.back() - nums[0], h = t;
        int sz = nums.size();
        for(int i = 0 ; i < sz - 1; ++i) h = min(h, nums[i + 1] - nums[i]); 
        pair<int, int> bd = {0, t};
        while(h <= t){
            int m = (t + h) >> 1, cnt = 0;
            // cerr << t << " " << h << "!\n";
            for(int i = 0; i < sz; ++i){
                int now = nums[i];
                int tmp = upper_bound(nums.begin(), nums.end(), now + m) - nums.begin() - i - 1;
                cnt += (tmp > 0? tmp : 0);
            }
            if(cnt < k){
                bd.first = max(bd.first, m);
                h = m + 1;
            }
            else if(cnt >= k){
                // cerr << cnt << " " << m << "\n";
                bd.second = min(bd.second, m);
                t = m - 1;
            }
            // else return m;
        }
        vector<int> ans;
        cerr << bd.first << " " << bd.second << "\n";
        int cnt = 0;
        if(bd.first == bd.second) return bd.first;
        for(int i = 0; i < sz; ++i){
            int now = nums[i];
            int tmp0 = upper_bound(nums.begin(), nums.end(), now + bd.first) - nums.begin() - 1;
            int tmp1 = upper_bound(nums.begin(), nums.end(), now + bd.second) - nums.begin() - 1;
            // cerr << tmp0 << " " << tmp1 << "\n";
            cnt += (tmp0 - i > 0? tmp0 - i: 0);
            for(int j = tmp0 + 1; j <= tmp1; ++j){
                ans.push_back(nums[j] - now);
                // cerr << nums[j] - now << " ";
            }
        }
        sort(ans.begin(), ans.end());
        // cerr << k - cnt - 1 << "\n";
        // if(cnt == k) return bd.first;
        return ans[k - cnt - 1];
    }
};
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