K Closest Points to Origin

Problem

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.

Intuition

The problem involves finding the k closest points to the origin in the Euclidean space. A natural approach is to calculate the distance of each point from the origin, store the distances in a min-heap, and then retrieve the k closest points from the heap.

Approach

Calculate Distances:

For each point [x, y] in the points array, calculate the distance from the origin using the formula dist = x^2 + y^2.
Create a min-heap, where each element is [dist, x, y].
Heapify and Retrieve:

Use heapq.heapify to convert the list of distances into a min-heap.
Retrieve the k closest points by repeatedly performing heapq.heappop operations on the min-heap.
Result:

The result is a list of k closest points, where each point is represented as [x, y].

Complexity

• Time complexity:

The time complexity is O(n log n), where n is the number of points. Calculating the distance for each point takes O(n), and heapifying the list takes O(n log n).

• Space complexity:

The space complexity is O(n) for the min-heap.

Code

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:
            dist, x, y = heapq.heappop(minheap)
            res.append([x, y])
            k -= 1
        
        return res