<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:
        c = Counter(nums)
        if len(c.keys()) <= 1000:
            return self.counterVersion(c, k)
        else:
            return self.iterDownVersion(nums, k)
    def iterDownVersion(self, nums, k):
        nums.sort()
        botb, topb = 0, nums[-1] - nums[0]
        while botb < topb:
            mid = (topb+botb) // 2
            val = self.numBelow(nums, mid)
            if val >= k:
                topb = mid
            else:
                botb = mid + 1
        return botb
    def numBelow(self, nums, dist):
        ret = 0
        left = 0
        for right in range(1, len(nums)):
            while nums[right] - nums[left] > dist:
                left += 1
            ret += right-left
        return ret
    def counterVersion(self, c, k):
        diffs = Counter()
        for a in c.keys():
            for b in c.keys():
                if a>b: continue
                if a==b:
                    diffs[0] += c[a]*(c[b]-1)/2
                else:
                    diffs[b-a] += c[a]*c[b]
        subk = k
        dkeys = list(diffs)
        dkeys.sort()
        i = 0
        while subk > diffs[dkeys[i]]:
            subk -= diffs[dkeys[i]]
            i += 1
        return dkeys[i]
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