<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>

struct FenwickTree {
    vector<int> bit; 
    int n;

    FenwickTree(int n) {
        this->n = n;
        bit.assign(n, 0);
    }

    int sum(int r) {
        int ret = 0;
        for (int idx = r; idx < n; idx = idx | (idx + 1)) {
            ret += bit[idx];
            // if (idx < 7) cout << idx << " " << bit[idx] << "\n";
        }
        // cout << ret << " @@\n";
        return ret;
    }

    int sum(int l, int r) {
        return sum(r) - sum(l - 1);
    }

    void add(int idx, int delta) {
        for (; idx >= 0; idx = (idx & (idx + 1)) - 1)
            bit[idx] += delta;
    }
};

class Solution {
public:
    int pairDistWithin(int d, vector<int>& nums) {
        FenwickTree ft(1e6 + 5);
        int res = 0;
        for (int i = 0; i < nums.size(); i++) {
            // cout << nums[i] << " " << ft.bit[1] << "\n";
            res += ft.sum(max(nums[i] - d, 0));
            // cout << "\n";
            ft.add(nums[i], 1);
        }
        return res;
    }
    int smallestDistancePair(vector<int>& nums, int k) {
        sort(nums.begin(), nums.end());
        int l = 0, r = 1e6, mid;
        while (true) {
            mid = (l + r) / 2;
            if (mid == l || mid == r)  break;
            if (pairDistWithin(mid, nums) >= k) {
                r = mid;
            } else {
                l = mid;
            }
        }
        cout << pairDistWithin(2, nums);
        for (mid = l; mid <= r; mid++) {
            if (pairDistWithin(mid, nums) >= k) return mid;
        }
        return 0;
    }
};
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