973 K Closest Points to Origin

• 973 K Closest Points to Origin
  • 문제
  • 조건
  • 예제
  • 해결
    • 1st
      • 1 생각
      • 1 코드
    • 2 리트코드 다른 풀이
      • 2 코드

973 K Closest Points to Origin

문제

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., $\sqrt{(x_{1} - x_{2})^{2}} + \sqrt{(y_{1} - y_{2})^{2}}$).
You may return the answer in any order. The answer is guaranteed to be unique (except for the order that it is in).

• 유클리디안 거리(Euclidean Distance) 개념과 구현해보기

조건

• $1 \le k \le points.length \le 10^{4}$
• $-10^{4} < x_{i}, y_{i} < 10^{4}$

예제

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]].

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.

해결

1st

1 생각

• 거리를 기준으로 정렬한다
• 거리가 키이고, 좌표가 값이 되어야 한다
• 거리를 기준으로 정렬하기 위해 최소 heap으로 구현된 heapq 사용

1 코드

def kClosest(self, points: List[List[int]], k: int) -> List[List[int]]:
    # 원점에서 K 번 가까운 점 목록을 순서대로 출력
    # 거리 순으로 정렬

    heap = []
    for x, y in points:
        # dist = math.sqrt((0 - x) ** 2 + (0 - y) ** 2)
        # 거리 계산을 모두 정확하게 할 필요는 없다
        dist = x ** 2 + y ** 2
        heapq.heappush(heap, (dist, x, y))
    
    result = []
    for _ in range(k):
        dist, x, y = heapq.heappop(heap)
        result.append([x, y])

    return result

2 리트코드 다른 풀이

2 코드

def kClosest2(self, points: List[List[int]], K: int) -> List[List[int]]:
    return sorted(points, key=lambda x: x[0]*x[0]+x[1]*x[1])[:K]


def kClosest3(self, points: List[List[int]], K: int) -> List[List[int]]:
    heap = []
    
    for (x, y) in points:
        dist = -(x*x + y*y)
        if len(heap) == K:
            heapq.heappushpop(heap, (dist, x, y))
        else:
            heapq.heappush(heap, (dist, x, y))
    
    return [(x,y) for (dist,x, y) in heap]