<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>

#define um unordered_map

class Solution {
public:
    // binary search + dp
    int smallestDistancePair(vector<int>& nums, int k) {
        sort(nums.begin(), nums.end());
        int maxDistance = nums.back() - nums[0];
        vector<int> prefixCount(2 * nums.back() + 1, 0); // number of elements smaller or equal to element i
        um<int, int> valueCount;
        for (auto& num : nums){
            valueCount[num]++;
        }
        prefixCount[0] = valueCount[0];
        for (int i = 1; i <= 2 * nums.back(); i++){
            prefixCount[i] = prefixCount[i-1] + valueCount[i];
        }
        int l = 0, r = maxDistance;
        while (l < r){
            int mid = l + (r - l) / 2;
            if (helper(nums, prefixCount, valueCount, mid) < k){
                l = mid+1;
            } else{
                r = mid;
            }
        }
        return l;
    }

    // count number of pairs smaller than or equal to the distance
    int helper(vector<int>& nums, vector<int>& prefixCount, um<int, int>& valueCount, int distance){
        int N = nums.size();
        int pairCount = 0;
        for (int i = 0; i < N; i++){
            if (i > 0 && nums[i] == nums[i-1]){
                continue;
            }
            int largerValueCount = prefixCount[nums[i] + distance] - prefixCount[nums[i]];
            int repeatPairCount = valueCount[nums[i]] * (valueCount[nums[i]] - 1) / 2;
            pairCount += largerValueCount * valueCount[nums[i]] + repeatPairCount;
        }
        return pairCount;
    }

};
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