<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) {
        int left = 0, right = points.size() - 1;

        while (left < right) {
            int p = partition(points, left, right);
            if (p == k) {
                break;
            } else if (p < k) {
                left = p + 1;
            } else {
                right = p - 1;
            }
        }

        std::vector<std::vector<int>> ans;
        for (int i = 0; i < k; ++i) {
            ans.push_back(points[i]);
        }
        return ans;
    }

private:
    int partition(std::vector<std::vector<int>>& points, int left, int right) {
        int p = left;
        std::vector<int> pivot = points[right];

        for (int i = left; i < right; ++i) {
            if (closer(points[i], pivot)) {
                std::swap(points[i], points[p++]);
            }
        }

        std::swap(points[right], points[p]);
        return p;
    }

    bool closer(const std::vector<int>& p1, const std::vector<int>& p2) {
        return p1[0] * p1[0] + p1[1] * p1[1] <= p2[0] * p2[0] + p2[1] * p2[1];
    }
};
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