<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:
    vector<vector<int>> kClosest(vector<vector<int>>& points, int k) {
        vector<vector<int>> res;
        auto comp = [](vector<int> p1, vector<int> p2){
            int d1 = pow(p1[0],2) + pow(p1[1],2);
            int d2 = pow(p2[0],2) + pow(p2[1],2);
            return (d1 < d2);
        };
        std::make_heap(res.begin(), res.end(), comp);
        for (int i=0; i<points.size(); i++){
            res.push_back(points[i]);
            std::push_heap(res.begin(), res.end(), comp);
            // for (int x=0; x<res.size(); x++){
            //     cout << res[x][0] << res[x][1] << " ";
            // }cout << endl;
            if(res.size() > k){
                std::pop_heap(res.begin(), res.end(), comp);
                res.pop_back();
            }
        }
        return res;
    }
};
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