<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) {
        // iteratively put element to priority queue (max heap) if size > k, take out the element.
        // store pair<int, int> , first is x^2+y^2, second is their pos
        auto compare = [](pair<int, int> num1, pair<int, int> num2)
        {
            return num1.first < num2.first;
        };
        priority_queue<pair<int, int>, vector<pair<int, int>>, decltype(compare)> maxHeap(compare);
        vector<vector<int>> result;

        for (int i = 0; i < points.size(); i++)
        {
            maxHeap.push({points[i][0] * points[i][0] + points[i][1] * points[i][1], i});
            if (maxHeap.size() > k)
                maxHeap.pop();
        }
        
        while (!maxHeap.empty())
        {
            int i = maxHeap.top().second;
            result.push_back(points[i]);
            maxHeap.pop();
        }
        return result;
    }
};
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