<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>

using ll = long long;
using vi = vector<ll>;

struct fenwick {
    int n;
    vi a;
    fenwick(int n) {
        this->n = n;
        a = vi(n);
    }

    void add(int p, ll x) {
        p++;
        while (p <= n) {
            a[p-1] += x;
            p += p & -p;
        }
    }

    ll sum(int p) {
        ll res = 0;
        while (p) {
            res += a[p-1];
            p -= p & -p;
        }
        return res;
    }

    ll sum(int l, int r) {
        return sum(r) - sum(l);
    }
};

class Solution {
public:
int smallestDistancePair(vector<int>& a, int k) {
        sort(a.begin(), a.end());
        ll mx = a.back();
        ll lo = 0, hi = mx, ans = -1;
        while (lo <= hi) {
            ll mid = (lo+hi)/2;
            fenwick fen(mx+1);
            ll cnt = 0;
            for (auto x : a) {
                ll lim = max(0ll, x - mid);
                cnt += fen.sum(lim,x+1);
                fen.add(x,1);
            }
            if (cnt >= k) {
                ans = mid;
                hi = mid - 1;
            }
            else lo = mid + 1;
        }
        return ans;
    }
};
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