<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>

from sortedcontainers import SortedList
class Solution:
    def smallestDistancePair(self, nums: List[int], k: int) -> int:
        """
        Binary search for the answer. How can you check how many pairs have distance <= X?
        sl.add(), and check largest and smallest gaps
        """

        
        # def check_gap(g):
        #     sl = SortedList()
        #     cur = 0
        #     for v in nums:
        #         ll = sl.bisect_left(v - g)                
        #         rr = sl.bisect_right(v + g)
        #         cur += rr - ll 
        #         print(ll, rr, cur)
        #         sl.add(v)
        #     return cur

        # print(0, check_gap(0))
        # print(1, check_gap(1))
        # print(2, check_gap(2))
        # print(3, check_gap(3))

        nums.sort()

        def check_gap(g):
            ans = 0 
            for i in range(len(nums)):
                v = nums[i]
                idx = bisect_right(nums, v + g)
                ans += (idx - i) - 1
            print(g, ans)
            return ans

        l, r = 0, max(nums) - min(nums) + 1
        while l < r:
            m = (l + r) >> 1
            f = check_gap(m) 
            if f >= k:
                r = m
            else:
                l = m + 1        
        return r

        

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