<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>

//窮舉的話是n!
//mantain一個list 
class Solution {
public:
    int smallestDistancePair(vector<int>& nums, int k) {
        sort(nums.begin(), nums.end());
        int maxElement = nums[nums.size()-1];
        map<int, int> value;
        //方便計算最後幾個數對的距離，實際最大距離為maxElement
        vector<int> prefixCount(maxElement*2);
        for(int i=0; i<nums.size();i++){
            ++value[nums[i]];
        }
        int index=0;
        for(int value=0; value<maxElement*2; value++){
            while(nums[index]<=value && index < nums.size())
                index++;
            prefixCount[value] = index;
        }
        int left = 0;
        int right = maxElement;
        while(left < right){
            int mid = (right + left) / 2;
            //cout << mid << endl;
            int result = countPair(value, prefixCount, mid, nums);
            //cout << result << " " << k << endl;
            if(result < k){
                left = mid+1;
            }else{
                right = mid;
            }
        }
        return left;
    }
    int countPair(map<int, int>value, vector<int> prefixCount, int maxDistance, const vector<int>& nums){
        int count = 0;
        for(int index=0; index<nums.size(); index++){
            
            int current = nums[index];
            int pairInRange = prefixCount[current+maxDistance] - prefixCount[current];
            int sameValue = (value[current] * (value[current] - 1))/2;
            count += (value[current] *  pairInRange)  + sameValue;
            while(index+1<nums.size() && nums[index] == nums[index+1])
                index++;
        }
        return count;
    }
}; 
//n!可以,push_back不行用hash的方式節省時間
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