프로그래머스 / 당구 연습

https://school.programmers.co.kr/learn/courses/30/lessons/169198

경계면을 기준으로 '접힌 상태를 편다'라고 생각하면 간단하다..
'피타고라스의 정의'를 이용해 거리를 계산.   

파이썬 

def solution(m, n, sx, sy, balls):
    return [
        min((sx - x) ** 2 + (sy - y) ** 2
            for x, y in ((-bx, by), (bx, -by), (2 * m - bx, by), (bx, 2 * n - by))
            if not ((sx == bx and (sy > by > y or sy < by < y))
                    or
                    (sy == by and (sx > bx > x or sx < bx < x)))
            )
        for bx, by in balls
    ]

이중 for 문을 이용한 리스트 컴프리헨션은 가독성이 좋진 않지만, 재미있는 코테를 위하여.. 

 

시작...

def solution(m, n, start_x, start_y, balls):
    answer = []
    for ball_x, ball_y in balls:
        temp = []
        for x, y in ((-ball_x, ball_y), (ball_x, -ball_y), (2 * m - ball_x, ball_y), (ball_x, 2 * n - ball_y)):
            if start_x == ball_x:
                if start_y > y:
                    if start_y > ball_y > y:
                        continue
                elif start_y < y:
                    if start_y < ball_y < y:
                        continue
            elif start_y == ball_y:
                if start_x > x:
                    if start_x > ball_x > x:
                        continue
                elif start_x < x:
                    if start_x < ball_x < x:
                        continue
            temp.append((start_x - x) ** 2 + (start_y - y) ** 2)
        answer.append(min(temp))
    return answer

줄여보자. 

def solution(m, n, start_x, start_y, balls):
    answer = []
    for ball_x, ball_y in balls:
        temp = []
        for x, y in ((-ball_x, ball_y), (ball_x, -ball_y), (2 * m - ball_x, ball_y), (ball_x, 2 * n - ball_y)):
            if start_x == ball_x:
                if start_y > ball_y > y or start_y < ball_y < y:
                    continue
            elif start_y == ball_y:
                if start_x > ball_x > x or start_x < ball_x < x:
                    continue
            temp.append((start_x - x) ** 2 + (start_y - y) ** 2)
        answer.append(min(temp))
    return answer

줄여보자. 

def solution(m, n, start_x, start_y, balls):
    answer = []
    for ball_x, ball_y in balls:
        temp = []
        for x, y in ((-ball_x, ball_y), (ball_x, -ball_y), (2 * m - ball_x, ball_y), (ball_x, 2 * n - ball_y)):
            if start_x == ball_x and (start_y > ball_y > y or start_y < ball_y < y):
                continue
            elif start_y == ball_y and (start_x > ball_x > x or start_x < ball_x < x):
                continue
            temp.append((start_x - x) ** 2 + (start_y - y) ** 2)
        answer.append(min(temp))
    return answer

줄여보자.. 

def solution(m, n, start_x, start_y, balls):
    answer = []
    for ball_x, ball_y in balls:
        temp = []
        for x, y in ((-ball_x, ball_y), (ball_x, -ball_y), (2 * m - ball_x, ball_y), (ball_x, 2 * n - ball_y)):
            if (start_x == ball_x and (start_y > ball_y > y or start_y < ball_y < y)) or \
                    (start_y == ball_y and (start_x > ball_x > x or start_x < ball_x < x)):
                continue
            temp.append((start_x - x) ** 2 + (start_y - y) ** 2)
        answer.append(min(temp))
    return answer

리스트 컴프리헨션으로 마무리하자. 

 

코틀린

import kotlin.math.*

class Solution {
    fun solution(m: Int, n: Int, sX: Int, sY: Int, balls: Array<IntArray>): IntArray {
        return balls.map { (bX, bY) ->
            arrayOf(-bX to bY, bX to -bY, 2 * m - bX to bY, bX to 2 * n - bY)
                .filter { (x, y) ->
                    !((sX == bX && (bY in y + 1 until sY || bY in sY + 1 until y))
                            || (sY == bY && (bX in (x + 1) until sX || bX in sX + 1 until x)))
                }.minOf { (x, y) ->
                    (sX - x.toDouble()).pow(2) + (sY - y.toDouble()).pow(2)
                }.toInt()
        }.toIntArray()
    }
}