<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, k):

        nums.sort()
        n = len(nums)

        max_elem = nums[-1]
        max_diff_range = max_elem + 1

        elems_count = [0] * max_diff_range
        value_count = [0] * (max_elem + 1)

        index = 0
        for i in range(max_diff_range):
            while index  < n and nums[index] <= i:
                index += 1
            
            elems_count[i] = index
        
        for i in range(n):
            value_count[nums[i]] += 1
        

        def count(mid):

            index, count = 0, 0
            
            while index < n:
                curr_v = nums[index]
                value_count_curr = value_count[curr_v]

                count_beyond_curr = elems_count[min(curr_v + mid, len(elems_count) - 1)] - elems_count[curr_v]

                count_duplicate_pairs = (value_count_curr * (value_count_curr - 1)) // 2

                count+= (count_beyond_curr * value_count_curr + count_duplicate_pairs)

                while index + 1 < n and nums[index] == nums[index+1]:
                    index+=1
                
                index+=1
            
            return count


        

        left, right = 0, max_elem

        while left < right:

            mid = (left + right) //2

            if count(mid) >= k:
                right = mid  
            else:
                left = mid + 1
            
        return left


       

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