973. K Closest Points to Origin

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

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

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

Example 1:

Input: points = [[1,3],[-2,2]], k = 1
Output: [[-2,2]]
Explanation:
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) < 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]].

Example 2:

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

Constraints:

• 1 <= k <= points.length <= 10<sup>4</sup>

• -10<sup>4</sup> <= x<sub>i</sub>, y<sub>i</sub> <= 10<sup>4</sup>

Solution:

class Solution:
    def kClosest(self, points: List[List[int]], k: int) -> List[List[int]]:
        minHeap = []
        for x, y in points:
            dist = (x ** 2) + (y ** 2)
            minHeap.append((dist, x, y))

        heapq.heapify(minHeap)
        res = []
        while k > 0:
            _, x, y = heapq.heappop(minHeap)
            res.append((x, y))
            k -= 1
        return res