1번

문제

이미 만들어져있는 아이디가 주어지고, 신규 고객이 아이디를 만들려할 때 이미 있는 아이디면 숫자를 하나씩 더 붙여 반환.

내 풀이

def solution(registered_list, new_id):
    if new_id not in registered_list:
        return new_id

    registered_list = set(registered_list)
    s_part, n_part = "", ""

    for i in range(len(new_id)):
        if not new_id[i].isnumeric():
            s_part += new_id[i]
        else:
            n_part += new_id[i]

    next_num = 1
    if n_part != "":
        next_num = int(n_part) + 1

    for num in range(next_num, 1000000):
        if s_part + str(num) not in registered_list:
            return s_part + str(num)
            
print(solution(["card", "ace13", "ace16", "banker", "ace17", "ace14"], "ace15") =="ace15")
print(solution(["cow", "cow1", "cow2", "cow3", "cow4", "cow9", "cow8", "cow7", "cow6", "cow5"], "cow") == "cow10")
print(solution(["bird99", "bird98", "bird101", "gotoxy"], "bird98") == "bird100")
print(solution(["apple1", "orange", "banana3"], "apple") =="apple")

처음에 리스트에서 in을 써서 시간초과났다. set으로 바꾸니 통과

2번

문제

한달동안 휴일이 주어지고, 연휴를 쓸 수 있는 날이 주어질 때, 가장 길게 쉬면 며칠인지 찾는 문제.

from itertools import combinations
import re
from collections import Counter
import copy

longest_day = 0

def count_longest(possible_combs, days):
    global longest_day
    days_copy = copy.deepcopy(days)

    for holiday in possible_combs:
        days_copy[holiday] = "HOL"

    cur_max = 0
    tmp_max = 0

    for i in range(30):
        if days_copy[i] in ["SAT", "SUN", "HOL"]:
            tmp_max += 1
        else:
            cur_max = max(cur_max, tmp_max)
            tmp_max = 0

    cur_max = max(cur_max, tmp_max)
    longest_day = max(longest_day, cur_max)


def solution(leave, day, holidays):
    global longest_day
    day_dict = {"MON": 0, "TUE" : 1, "WED" : 2, "THU" : 3, "FRI" : 4, "SAT" : 5, "SUN" : 6}
    day_list = ["MON", "TUE", "WED", "THU", "FRI", "SAT", "SUN"]

    cur_day_index = day_dict[day]
    days = [""] * 30

    for i in range(30):
        days[i] = day_list[cur_day_index]
        cur_day_index = (cur_day_index + 1) % 7

    for holiday in holidays:
        holiday -= 1
        days[holiday] = "HOL"

    work_days = []
    for i in range(30):
        if days[i] not in ["SAT", "SUN", "HOL"]:
            work_days.append(i)

    if len(work_days) <= leave:
        return 30

    for possible_combs in list(combinations(work_days, leave)):
        count_longest(possible_combs, days)

    return longest_day

#print(solution(4, "FRI", [6, 21, 23, 27, 28]) == 10)
#print(solution(3, "SUN", [2, 6, 17, 29]) == 8)
print(solution(30, "MON", [1, 2, 3, 4, 28, 29, 30]) == 30)

3번

문제

격자 무니 게임판에서 위에서 아이템이 떨어지고, 3개 이상 뭉쳐있으면 터져서 없어지고 위에것들이 내려오는 게임 구현

내 풀이

from collections import defaultdict, deque
from itertools import combinations
import re
from collections import Counter
import copy

dx = [-1, 0, 0, 1]
dy = [0, -1, 1, 0]

def print_board(board):
    for i in range(len(board)):
        for j in range(len(board[0])):
            print(board[i][j], end=" ")
        print()


def bfs(visited, board, i ,j):
    visited[i][j] = True
    queue = deque()
    queue.append([i, j])

    tmp_visited_list = []

    while queue:
        y, x = queue.popleft()
        tmp_visited_list.append([y, x])

        for k in range(4):
            ny, nx = y + dy[k], x + dx[k]
            if 0 <= ny < len(board) and 0 <= nx < len(board[0]):
                if not visited[ny][nx] and board[ny][nx] == board[i][j]:
                    visited[ny][nx] = True
                    queue.append([ny, nx])

    return tmp_visited_list


def bomb_board(board, stacks):
    visited = [[False] * len(board[0]) for _ in range(len(board))]
    for i in range(len(board)):
        for j in range(len(board[0])):
            if not visited[i][j] and board[i][j] != 0:
                tmp_visited_list = bfs(visited, board, i, j)

                if len(tmp_visited_list) >= 3:
                    for y, x in tmp_visited_list:
                        board[y][x] = -1


def change_board(board, stacks):
    board = [[0] * 6 for _ in range(6)]
    for j in range(6):
        i = 5
        stack_copy = copy.deepcopy(stacks[j])

        while stack_copy:
            board[i][j] = stack_copy.pop(0)
            i -= 1

    bomb_board(board, stacks)

    return board


def change_stack(board):
    stacks = [[] for _ in range(6)]
    for j in range(6):
        for i in range(5, -1, -1):
            if board[i][j] not in [0, -1]:
                stacks[j].append(board[i][j])
    return stacks


def is_bombed(board):
    for i in range(len(board)):
        for j in range(len(board[0])):
            if board[i][j] == -1:
                return True
    return False


def solution(macaron):
    answer = []
    board = [[0] * 6 for _ in range(6)]
    stacks = [[] for _ in range(6)]
    bombed = False

    for mac in macaron: #  [떨어뜨린 위치, 마카롱의 색]
        pos, color = mac[0] -1, mac[1]
        stacks[pos].append(color)
        board= change_board(board, stacks)
        bombed = is_bombed(board)
        stacks = change_stack(board)

        while bombed:
            board = change_board(board, stacks)
            stacks = change_stack(board)
            bombed = is_bombed(board)

    board = change_board(board, stacks)
    for i in range(6):
        answer.append(''.join(map(str, board[i])))
    return answer


print(solution([[1,1],[2,1],[1,2],[3,3],[6,4],[3,1],[3,3],[3,3],[3,4],[2,1]]) == ["000000","000000","000000","000000","000000","204004"])
print(solution([[1,1],[1,2],[1,4],[2,1],[2,2],[2,3],[3,4],[3,1],[3,2],[3,3],[3,4],[4,4],[4,3],[5,4],[6,1]]) == ["000000","000000","000000","000000","000000","404001"])

4번

SELECT DISTINCT(subway_stations.id) AS "ID", subway_stations.name as "NAME"
FROM subway_stations left join line_routes on subway_stations.id = line_routes.station_id
WHERE subway_stations.id not in (SELECT station_id FROM line_routes WHERE line_color = "Red" or line_color = "Green")
ORDER BY subway_stations.id