Given an array of points where points[i] = [xi, yi] 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., √(x1 - x2)2 + (y1 - y2)2).
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.
1 <= k <= points.length <= 104-104 <= xi, yi <= 104
import heapq
class Solution:
def kClosest(self, points: list[list[int]], k: int) -> list[list[int]]:
heap = []
for point in points:
distance = -(point[0] ** 2 + point[1] ** 2) # Negate to create a max heap
if len(heap) < k:
heapq.heappush(heap, (distance, point))
else:
heapq.heappushpop(heap, (distance, point))
return [point for _, point in heap]Just use libraries instead of writing one yourself.
O(n log k), where n is the number of points in the input array. We insert each point into the heap, which takes O(log k) time, and we do this for all n points.
O(k) for the max heap, where k is the number of closest points we want to find.