문제 바로가기

from itertools import combinations

def solution(relation):
    answer = 0
    
    # ls : [0,1] 과 같은 index 리스트
    def unique(ls):
        res = []
        for re in relation:
            k = ""
            for i in ls:
                k += re[i] + " "
            res.append(k)
        sres = set(res)
        return len(res) == len(sres)
    
    def check(k, ls):
        for l in ls:
            if len(set(l) - set(k)) == 0:
                return False
        return True
    
    ls = []
    for i in range(1, len(relation[0]) + 1):
        for k in combinations([i for i in range(len(relation[0]))], i):
            if unique(k) and check(k, ls) :
                answer += 1
                ls.append(k)
                
    return answer

improve

res = [tuple([item[key] for key in ls]) for item in relation]
이런 식으로 할 수도 있음

부분집합인지 확인하는 함수

a.issubset(b)
b.issupperset(a)

나는 len(set(a) - set(b)) == 0 을 이용했다.