<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 FenwickTree {
    vector<int> fenTree, nums;
public:
    FenwickTree(int n, vector<int> &arr) {
        this->nums = arr;
        fenTree.resize(1e7+5);
        for(int i=1;i<=n;i++) {
            addValue(arr[i-1]+1, 1);
        }
    }
    void addValue(int ind, int value) {
        while(ind < fenTree.size()) {
            fenTree[ind] += value;
            ind += (ind & (-ind));
        }
    }
    int getSum(int ind) {
        int sum=0;
        while(ind > 0) {
            sum += fenTree[ind];
            ind -= (ind & (-ind));
        }
        return sum;
    }
    int getRangeSum(int left,int right) {
        return getSum(right+1) - getSum(left);
    }
};
class Solution {
    bool isValid(int curDiff, vector<int> &nums, int k) {
        int n=nums.size();
        FenwickTree fenTree(n, nums);
        int count = 0;
        for(int i=0;i<n;i++) {
            fenTree.addValue(nums[i]+1, -1);
            int left = nums[i]-curDiff, right = nums[i]+curDiff;
            count += fenTree.getRangeSum(left, right);
        }
        // cout<<curDiff<<' '<<count<<endl;
        return count >= k;
    }
public:
    int smallestDistancePair(vector<int>& nums, int k) {
        int n=nums.size();
        int low = 0;
        int high = *max_element(nums.begin(),nums.end()) - *min_element(nums.begin(),nums.end());
        int ans = -1;
        while(low <= high) {
            int mid = low+(high-low)/2;
            if(isValid(mid, nums, k)) {
                ans = mid;
                high = mid-1;
            } else {
                low = mid+1;
            }
        }

        return ans;
    }
};
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