KAKAOBLINDRECRUITMENT 3

[프로그래머스 Lv2 - 2019 KAKAO BLIND RECRUITMENT] 오픈채팅방 (Python)

def solution(record): li = dict() answer = [] sen = [] for i, rec in enumerate(record) : rec = rec.split() op = rec[0] id = rec[1] if op == "Enter" : name = rec[2] if id in li.keys() : if li[id] != name : idx = [i for i, x in enumerate(sen) if x[0] == id] for i in idx : sen[i][1] = name li[id] = name sen.append([id, name, 1]) elif op == 'Leave' : name = li[id] sen.append([id, name, 2]) else : na..

알고리즘 2022.07.02

[프로그래머스 Lv1 - 2019 KAKAO BLIND RECRUITMENT] 실패율 (Python)

def solution(N, stages): length = len(stages) answer = [] fail = dict() for i in range(1, N+1) : cnt = stages.count(i) val = cnt / length fail[i] = val length -= cnt fail = sorted(fail.items(), key=lambda x:x[1], reverse=True) for i in fail : answer.append(i[0]) return answer ㅎㅎ.. 웬일로 쉽게 끝나나 했다 ,, def solution(N, stages): length = len(stages) answer = [] fail = dict() for i in range(1, N+1) : cn..

알고리즘 2022.06.30