<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 countLEPair(vector<int> const &valPrefix,
                    unordered_map<int,int> &valCount,
                    vector<int> const &nums,
                    int target) {
        // count # of pairs that is lesser or equal to target
        int total = 0;
        const int N = nums.size();
        int index = 0;
        while (index < N) {
            int cur = nums[index], curCount = valCount[cur];
            total += (valPrefix[cur+target]-valPrefix[cur]) * curCount;
            total += (curCount*(curCount-1))/2;
            index += curCount;
        }
        return total;
    }
    int smallestDistancePair(vector<int>& nums, int k) {
        sort(nums.begin(), nums.end());
        const int N = nums.size();
        int mxDist = nums.back() - nums.front();
        vector<int> valPrefix(mxDist*2+1);
        unordered_map<int,int> valCount;
        int index = 0;
        for (int val = 0; val < valPrefix.size(); val++) {
            while (index< N && nums[index] <= val) ++index;
            valPrefix[val] = index;
        }
        for (auto e: nums) {
            ++valCount[e];
        }
        int l = 0 , r = mxDist;
        while (l<r) {
            int m = l+(r-l)/2;
            auto count = countLEPair(valPrefix, valCount, nums, m);
            if (count < k) {
                l = m+1;
            } else {
                r = m;
            }
        }
        return l;
    }
};
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