<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:
  def smallestDistancePair(self, nums: List[int], k: int) -> int:
    # binary search with dynamic programming

    n = len(nums)
    nums.sort()

    max_d = nums[-1]*2
    pre = [0] * max_d
    val = defaultdict(int)

    def count(d):
      c, i = 0, 0
      while i < n:
        cur = nums[i]
        cur_cnt = val[cur]
        gt = pre[min(cur + d, len(pre) - 1)] - pre[cur]
        eq = cur_cnt * (cur_cnt - 1) // 2
        c += gt * cur_cnt + eq
        while i + 1 < n and nums[i] == nums[i+1]:
          i += 1
        i += 1

      return c

    idx = 0
    for i in range(max_d):
      while idx < n and nums[idx] <= i:
        idx += 1
      pre[i] = idx
    for i in range(n):
      val[nums[i]] += 1

    l, h = 0, nums[-1] - nums[0]
    while l < h:
      m = l + (h - l) // 2
      cnt = count(m)
      if cnt < k:
        l = m + 1
      else:
        h = m

    return l

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