<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 {
public:
    int smallestDistancePair(vector<int>& nums, int k) {
        vector<int> pref(1e6 + 1); 
        vector<int> differences(1e6 + 2); 
        int n = nums.size(); 
        for (int i = 0; i < n; i ++) { 
            for (int j = i + 1; j < n; j ++) { 
                pref[abs(nums[i] - nums[j])] ++;  
                differences[abs(nums[i] - nums[j])] ++;  
            }
        }
        
        for (int i = 1; i <= 1e6; i ++) { 
            pref[i] += pref[i - 1]; 
        }
        auto valid = [&](int x) { 
            return pref[x] >= k; 
        };

        int l = 0, r = 1e6, ans = 1e6 + 1; 
        while (l <= r) { 
            int mid = (l + r) / 2; 
            if (valid(mid)) { 
                ans = mid; 
                r = mid - 1; 
            } else { 
                l = mid + 1; 
            }
        }
        return ans;  
    }
};
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