[프로그래머스 Lv2, python] - 지게차와 크레인

https://school.programmers.co.kr/learn/courses/30/lessons/388353

프로그래머스

def fork(storage, box):  # 지게차
    dx, dy = [0, 0, 1, -1], [1, -1, 0, 0]
    index = []

    for i in range(1, len(storage) - 1):
        for j in range(1, len(storage[0]) - 1):
            if storage[i][j] == box:
                for k in range(4):
                    ni, nj = i + dx[k], j + dy[k]
                    if storage[ni][nj] == "0":
                        index.append((i, j))
                        break

    for i, j in index:
        storage[i][j] = "0"
        spread_outside(storage, i, j)

def crane(storage, box):  # 크레인
    for i in range(1, len(storage) - 1):
        for j in range(1, len(storage[0]) - 1):
            if storage[i][j] == box:
                storage[i][j] = "1"
                spread_outside(storage, i, j)

def spread_outside(storage, x, y):
    dx, dy = [0, 0, 1, -1], [1, -1, 0, 0]
    if any(storage[x + dx[k]][y + dy[k]] == "0" for k in range(4)): # 외부와 이어지는가?
        storage[x][y] = "0"
        for k in range(4):
            nx, ny = x + dx[k], y + dy[k] 
            if storage[nx][ny] == "1": # 0으로 만든 뒤 다른것들도 외부와 이어지는지?
                spread_outside(storage, nx, ny)

def solution(storage, requests):
    answer = 0
    storage = [list("0" + row + "0") for row in storage] # 외부를 0으로 만들어주기(둘러싸도록)
    storage.insert(0, list("0" * len(storage[0])))
    storage.append(list("0" * len(storage[0])))

    for req in requests:
        if len(req) == 1:
            fork(storage, req)
        else:
            crane(storage, req[0])

    for i in range(1, len(storage) - 1):
        for j in range(1, len(storage[0]) - 1):
            if storage[i][j] not in ["0", "1"]:
                answer += 1
    return answer

 

 

# 처음 풀이. 정답률 25퍼,, 
import copy

def check(storage, r, c): 
    dic = [[0, -1], [0, 1], [1, 0], [-1, 0]] # 좌 우 상 하
    for x, y in dic: # 상하좌우 중 빈 문자열 있는지?
        dx, dy = x + r, y + c
        if 0 <= dx <= len(storage) and 0 <= dy <= len(storage[0]):
            if storage[dx][dy] == "0":
                return 1
    return 0

def change(storage): # 리스트로 각각 나눠주기, 0을 외부로 채워주기
    new = [["0"] * (len(storage[0])+1)]
    for i in storage:
        tmp = ["0"]
        for j in i:
            tmp.append(j)
        tmp.append("0")
        new.append(tmp)
    new.append(["0"] * (len(storage[0])+1))
    return new

def solution(storage, requests):
    ans = 0
    new = change(storage)
    new_2 = copy.deepcopy(new)
    
    for alp in requests:
        le = len(alp)
        if le == 2:
            for i in range(len(new)):
                for j in range(len(new[i])):
                    if new[i][j] == alp[0]:
                        if(check(new, i, j)): 
                            new[i][j] ='0'  
                        else:
                            new[i][j] ='1'
                        ans += 1
        else:
            for i in range(len(new)):
                for j in range(len(new[i])):
                    if new[i][j] == alp:
                        if(check(new_2, i, j)): 
                            new[i][j] ='0'
                            ans += 1
        new_2 = copy.deepcopy(new)
    return len(storage) * len(storage[0]) - ans