<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:
    const int offset = 1e6;
    const int N = 2 * 1e6 + 1;
    int smallestDistancePair(vector<int>& nums, int k) {
        int n = nums.size();
        //segment tree
        vector<int> t(2 * N, 0);
        build(nums, t);
        int l = 0, r = 1e6 + 1;
        while(l < r){
            int mid = (l + r) >> 1;
            int cnt = 0;
            for(int i = 0; i < n; i ++){
                int cur = nums[i];
                modify(t, cur + offset, -1);
                cnt += query(t, cur + offset - mid, min(cur + offset + mid + 1, N - 1));
            }
            for(int i = 0; i < n; i ++){
                int cur = nums[i];
                modify(t, cur + offset, 1);
            }
            if(cnt < k){
                l = mid + 1;
            }
            else{
                r = mid;
            }
        }
        
        return l;
    }
    
    void build(vector<int>& nums, vector<int>& t){
        for(int num : nums){
            int v = num + offset;
            t[v + N] ++;
        }
        for(int i = N - 1; 0 < i; -- i){
            t[i] = t[i << 1] + t[i << 1 | 1];
        }
    }
    
    void modify(vector<int>& t, int p, int diff){
        for(t[p += N] += diff; p > 1; p >>= 1){
            t[p >> 1] = t[p] + t[p ^ 1];
        }
    }
    
    int query(vector<int>& t, int l, int r){
        //sum of [l, r)
        int ret = 0;
        for(l += N, r += N; l < r; l >>= 1, r >>= 1){
            if(l & 1){
                ret += t[l ++];
            }
            if(r & 1){
                ret += t[-- r];
            }
        }
        
        return ret;
    }
};
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