973. K Closest Points to Origin (Python)

Related Topic

Heap.

Description

We have a list of points on the plane. Find the K closest points to the origin (0, 0).

(Here, the distance between two points on a plane is the Euclidean distance.)

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

Sample I/O

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]]
(The answer [[-2,4],[3,3]] would also be accepted.)

Note

• 1 <= K <= points.length <= 10000
• -10000 < points[i][0] < 10000
• -10000 < points[i][1] < 10000

Methodology

Use max heap to maintain the first Kth largest pair. The largest pair is calculated as sqrt(x*x+y*y). However you don’t need to calculate sqrt, this won’t affect the result and save a little time.

def kClosest(self, points: List[List[int]], K: int) -> List[List[int]]:
        heap = []
        for x,y in points:
            if len(heap)<K:
                heapq.heappush(heap,[-(x*x+y*y),[x,y]])
            else:
                heapq.heappushpop(heap,[-(x*x+y*y),[x,y]])
        return [pair for value, pair in heap]

BigO

Time complexity : O(nlogn) where n is the size of input.