Dev On The Coffee Dev On The Coffee

# 문제 내 풀이 # 내 풀이 - 시간 복잡도 O(N²) def solution(phone_book): for i in range(len(phone_book)): for j in range(i + 1, len(phone_book)): min_len = min(len(phone_book[i]), len(phone_book[j])) if phone_book[j][:min_len] == phone_book[i][:min_len]: return False return True 지향할 답안 # 모범 답안 - 시간 복잡도 O(NlogN) def solution(phoneBook): phoneBook = sorted(phoneBook) for p1, p2 in zip(phoneBook, phoneBook[1:]): ..