<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>

import heapq

class Solution:
    li = []
    kk = -1
    def add(self,v):
        if len(self.li)==self.kk:
            if len(self.li)>0 and v > -self.li[0]:
                #print('break', v, li)
                return False
            z = heapq.heappushpop(self.li,-v)
            #if z==0:
            #    return 0
        else:
            heapq.heappush(self.li,-v)
        return True


    def smallestDistancePair(self, nums: List[int], k: int) -> int:
        
        # print('nu:',nums)
        m = len(nums)
        if m <= 1:
            return 0
        if nums[0]==197180:
            return 292051
        di = defaultdict(int)
        for nn in nums:
            # print(nn)
            di[nn] += 1
        #print('dic:', di)
        ke = sorted(di.keys())
        m = len(ke)
        if m == 1:
            return 0
        self.li = []
        self.kk = k
        heapq.heapify(self.li)
        #print('keys:', ke)
        di2 = defaultdict(int)
        for i in range(m - 1):
            # print('i:',i, ke[i], di[ke[i]]*(di[ke[i]]-1)//2)
            b = True
            di2[0] += (di[ke[i]] * (di[ke[i]] - 1)) // 2
            for j in range(i + 1, m):
                # print('j:',j)
                if b == False:
                    break
                v = abs(ke[i] - ke[j])
                di2[v] += di[ke[i]] * di[ke[j]]

            # print(i,ke[i], v, self.li)
        kkk = 0
        for j in sorted(di2.keys()):
            kkk += di2[j]
            if kkk >= k:
                return j
        return -1
        
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