<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 count_pairs(self, nums, prefix_count, value_count, max_distance):
        count = 0
        n = len(nums)
        idx = 0
        while idx < n:
            curr = nums[idx]
            value_count_for_curr = value_count[curr]
            pairs_with_larger_values = (prefix_count[min(curr+max_distance, len(prefix_count)-1)] - prefix_count[curr])
            pairs_with_same_values = (value_count_for_curr * (value_count_for_curr -1)//2)

            count += (pairs_with_larger_values * value_count_for_curr + pairs_with_same_values)
            while idx + 1 < n and nums[idx] == nums[idx+1]:
                idx += 1
            idx += 1
        return count
    def smallestDistancePair(self, nums: List[int], k: int) -> int:
        nums.sort()
        n = len(nums)

        max_element = nums[-1]
        max_possible_distance = max_element * 2

        prefix_count = [0] * max_possible_distance
        value_count = [0] * (max_element + 1)

        idx = 0
        for value in range(max_possible_distance):
            while idx < n and nums[idx] <= value:
                idx += 1
            prefix_count[value] = idx
        for i in range(n):
            value_count[nums[i]] += 1

        lo, hi = 0, max_element
        while lo<hi:
            mid = (lo+hi)//2
            count = self.count_pairs(nums, prefix_count, value_count, mid)
            if count < k:
                lo = mid + 1
            else:
                hi = mid
        return lo
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