[1차] 셔틀버스

from collections import deque BUSSTART = 540 CREW, BUS = 0, 1 def h2m(t): return 60*int(t.split(':')[0]) + int(t.split(':')[1]) def m2h(t): return f'{t//60:02}:{t%60:02}' def solution(n, t, m, timetable): crewtimes = [(h2m(t), CREW) for t in timetable] bustimes = [(BUSSTART + t*i, BUS) for i in range(n)] timepoints = crewtimes + bustimes timepoints.sort() q, groups = deque(), [] for t, typ in timepoints: if typ == CREW: q.append(t) else: group = tuple(q.popleft() for _ in range(min(m, len(q)))) groups.append((group, t)) if len(groups[-1][0]) < m: return m2h(groups[-1][1]) # 막차에 꽉 찼을 경우입니다. else: # 막차에 가장 늦게 탈 수 있는 경우를 찾습니다. last, i = groups[-1][0][-1], m-1 while i > 0: if groups[-1][0][i-1] != last: break i -= 1 return m2h(groups[-1][0][i] - 1) answer = '' return answer