[LeetCode] 1779. Find Nearest Point That Has the Same X or Y Coordinate

- Problem

1779. Find Nearest Point That Has the Same X or Y Coordinate

You are given two integers, x and y, which represent your current location on a Cartesian grid: (x, y). You are also given an array points where each points[i] = [ai, bi] represents that a point exists at (ai, bi). A point is valid if it shares the same x-coordinate or the same y-coordinate as your location.

Return the index (0-indexed) of the valid point with the smallest Manhattan distance from your current location. If there are multiple, return the valid point with the smallest index. If there are no valid points, return -1.

The Manhattan distance between two points (x1, y1) and (x2, y2) is abs(x1 - x2) + abs(y1 - y2).

Example 1:

Input: x = 3, y = 4, points = [[1,2],[3,1],[2,4],[2,3],[4,4]]
Output: 2
Explanation: Of all the points, only [3,1], [2,4] and [4,4] are valid. Of the valid points, [2,4] and [4,4] have the smallest Manhattan distance from your current location, with a distance of 1. [2,4] has the smallest index, so return 2.

Example 2:

Input: x = 3, y = 4, points = [[3,4]]
Output: 0
Explanation: The answer is allowed to be on the same location as your current location.

Example 3:

Input: x = 3, y = 4, points = [[2,3]]
Output: -1
Explanation: There are no valid points.

Constraints:

• 1 <= points.length <= 10^4
• points[i].length == 2
• 1 <= x, y, ai, bi <= 10^4

정수 x, y, 좌표를 원소로 갖는 이차원 배열 points가 주어진다.

point를 배열 points의 임의의 원소라고 가정하자.
point[0]==x 혹은 point[1]==y를 만족하는 point를 찾고 이를 주어진 정수 [x, y]와의 최소 맨하탄 거리를 갖는 point의 인덱스를 찾는 문제이다.

맨하탄 거리는 다음과 같이 정의한다.

The Manhattan distance between two points (x1, y1) and (x2, y2) is abs(x1 - x2) + abs(y1 - y2).

---

- 내 풀이

class Solution:
    def nearestValidPoint(self, x: int, y: int, points: List[List[int]]) -> int:
        answer, min_distance = -1, float('inf')
        
        for i, (coor_x, coor_y) in enumerate(points):
            if coor_x == x or coor_y == y:
                curr_distance = abs(coor_x - x) + abs(coor_y - y)
                if curr_distance < min_distance:
                    min_distance = curr_distance
                    answer = i
        
        return answer

- 결과