<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 checkCount(unordered_map<int, int> freq, vector<int> count, int dist) {
        int numCount = 0;
        for (auto& pair : freq) {
            int val = pair.first;
            if (val + dist < count.size()) {
                numCount += (count[val + dist] - count[val]) * pair.second;
            }
            else {
                numCount += (count[count.size() - 1] - count[val]) * pair.second;
            }
            numCount += pair.second * (pair.second - 1) / 2;
        }
        return numCount;
    }

    int smallestDistancePair(vector<int>& nums, int k) {
        sort(nums.begin(), nums.end());

        int low = 0, high = nums[nums.size() - 1] - nums[0];

        //i want an array with how many numbers are it/greater than it
        //i want a map of frequency of number
        unordered_map<int, int> freq;
        vector<int> count(nums[nums.size() - 1] + 1, 0);

        for (int i = 0; i < nums.size(); i++) {
            int val = nums[i];
            count[val]++;
            freq[val]++;
        }

        for (int i = 1; i < count.size(); i++) {
            count[i] += count[i - 1];
        }

        // for (int i = 0; i < count.size(); i++) {
        //     cout << count[i] << " ";
        // }
        // cout << endl;
        // for (auto& pair : freq) {
        //     cout << pair.first << " " << pair.second << endl;
        // }

        while (low < high) {
            int mid = low + (high - low) / 2;
            int amount = checkCount(freq, count, mid);
            cout << mid << " " << amount << endl;

            if (amount < k) {
                low = mid + 1;
            }
            else {
                high = mid;
            }
        }
        return low;
    }
    
};
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