<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:
        nums.sort()
        self.nums = nums
        n = len(self.nums)
        return self.bSearch(-1, self.nums[-1] - self.nums[0], 0, n * (n - 1) / 2, k)
    
    def bSearch(self, a: int, b: int, A: int, B: int, k: int) -> int:
        # B pairs <= b and A pairs <= a
        # k paris <= b
        # k pairs <= a
        # a is the answer in that case

        # k pairs <= b
        # k - 1 pairs <= b -1

        if A == B:
            return a
        if a == b - 1:
            return b
        c = (a + b) // 2
        # count the number of pairs <= c
        cond = []
        C = 0
        for num in self.nums:
            while cond and num > cond[0]:
                cond.pop(0)
            C += len(cond)
            cond.append(num + c)
        if C < k:
            return self.bSearch(c, b, C, B, k)
        else:
            return self.bSearch(a, c, A, C, k)
18658