각 a번 수포자가 찍을 때 다른데
- 1번은 1~5 반복
- 2번은 [2 1][2 3] [2 4][2 5] [2 1][2 3] 이렇게, 2는 제외!
- 3번은 3-1-2-4-5를 각각 두 번씩
브루트포스->answer와 하나씩 매칭(==)시켜서 cnt가 많은 사람이 누구인지
높은 점수 받은 사람이 여러 명이면, asc정렬
def solution(answers):
answer = [0,0,0]
leng = len(answers)
ans = []
man1 = [i+1 for i in range(5)]*2000
man2 = [2,1,2,3,2,4,2,5] * 1250
man3 = [3,3,1,1,2,2,4,4,5,5]*1000
for i in range(leng):
if man1[i] == answers[i]: answer[0] += 1
if man2[i] == answers[i]: answer[1] += 1
if man3[i] == answers[i]: answer[2] += 1
indexer = answer.index(max(answer))
for i in range(3):
if answer[i] == answer[indexer]:
ans.append(i+1)
return ans
print(solution([1,2,3,4,5]))
print(solution([1,3,2,4,2]))
def solution(answers):
first = [1,2,3,4,5]
second = [2,1,2,3,2,4,2,5]
third = [3,3,1,1,2,2,4,4,5,5]
ret = [0,0,0]
for i in range(len(answers)):
if answers[i] == first[i%5]:
ret[0] += 1
if answers[i] == second[i%8]:
ret[1] += 1
if answers[i] == third[i%10]:
ret[2] += 1
maxval = max(ret)
answer = []
for i in range(len(ret)):
if ret[i] == maxval:
answer.append(i+1)
return answer