K Closest Points to Origin
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0900-0999/0973.K%20Closest%20Points%20to%20Origin/images/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>
Geometry; Array; Math; Divide and Conquer; Quickselect; Sorting; Heap (Priority Queue)
func kClosest(points [][]int, k int) [][]int {
	sort.Slice(points, func(i, j int) bool {
		return math.Hypot(float64(points[i][0]), float64(points[i][1])) < math.Hypot(float64(points[j][0]), float64(points[j][1]))
	})
	return points[:k]
}
