<p>Given an array of <code>points</code> where <code>points[i] = [x<sub>i</sub>, y<sub>i</sub>]</code> represents a point on the <strong>X-Y</strong> plane and an integer <code>k</code>, return the <code>k</code> closest points to the origin <code>(0, 0)</code>.</p>

<p>The distance between two points on the <strong>X-Y</strong> plane is the Euclidean distance (i.e., <code>&radic;(x<sub>1</sub> - x<sub>2</sub>)<sup>2</sup> + (y<sub>1</sub> - y<sub>2</sub>)<sup>2</sup></code>).</p>

<p>You may return the answer in <strong>any order</strong>. The answer is <strong>guaranteed</strong> to be <strong>unique</strong> (except for the order that it is in).</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/03/closestplane1.jpg" style="width: 400px; height: 400px;" />
<pre>
<strong>Input:</strong> points = [[1,3],[-2,2]], k = 1
<strong>Output:</strong> [[-2,2]]
<strong>Explanation:</strong>
The distance between (1, 3) and the origin is sqrt(10).
The distance between (-2, 2) and the origin is sqrt(8).
Since sqrt(8) &lt; sqrt(10), (-2, 2) is closer to the origin.
We only want the closest k = 1 points from the origin, so the answer is just [[-2,2]].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> points = [[3,3],[5,-1],[-2,4]], k = 2
<strong>Output:</strong> [[3,3],[-2,4]]
<strong>Explanation:</strong> The answer [[-2,4],[3,3]] would also be accepted.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= k &lt;= points.length &lt;= 10<sup>4</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= x<sub>i</sub>, y<sub>i</sub> &lt;= 10<sup>4</sup></code></li>
</ul>


class Solution {
public:
    double solve(vector<int>points){
        int a=points[0];
        int b=points[1];
        double formula=sqrt((pow(a,2)-0)+(pow(b,2)-0));
        return formula;
    }
    vector<vector<int>> kClosest(vector<vector<int>>& points, int k) {
        //replace int with pair<double,vector<int>> the type 
        //rest same as q<int,vector<int>,greater<int>>
        priority_queue<pair<double,vector<int>> ,vector<pair<double,vector<int>>> ,greater<pair<double,vector<int>>> >q;
        for(int i=0;i<points.size();i++){
            double ans=solve(points[i]);
            q.push({ans,points[i]});    
        }

        vector<vector<int>>result;
        while(k--){
            if(q.size()>0){
                auto temp=q.top();
                q.pop();
                result.push_back(temp.second);
            }
        }
        return result;
        
        
        
    }
};
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