<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 search(vector<int>& tree, int i) {
        if (i < 1) {
            return 0;
        }
        int ans = 0;
        while (i > 0) {
            ans += tree[i];
            i -= (i & -i);
        }
        return ans;
    }
    void update(vector<int>& tree, int i) {
        while (i < tree.size()) {
            tree[i]++;
            i += (i & -i);
        }
    }
    int findPairs(vector<int>& nums, int mid, int limit) {
        vector<int> tree(limit + 1);
        int ans = 0;
        for (auto &c: nums) {
            ans += search(tree, c - mid);
            update(tree, c);
        }
        return ans;
    }
    int smallestDistancePair(vector<int>& nums, int k) {
        sort(nums.begin(), nums.end());
        for (auto &c: nums) {
            c++;
        }
        int low = 0;
        int high = nums.back() - nums[0];
        int numAccom = (nums.size() * (nums.size() - 1))/2 - (k - 1);
        while (low <= high) {
            int mid = (low + high)/2;
            // cout << low << " " << mid << " " << high << endl;
            int temp = findPairs(nums, mid, nums.back());
            // cout << temp << endl;
            if (temp < numAccom) { // accomodating less, then decrease distance
                high = mid - 1;
            }
            else {
                low = mid + 1; // accomodating a lot, increase distance
            }
        }
        return high;
        
    }
};
218696