def solution(n, words):
    pos = [0 for _ in range(n)]
    pos[0] = 1
    lst = set([words[0]])
    for i in range(1, len(words)):
        pos[i % n] += 1
        if words[i-1][-1] != words[i][0]:
            return (i % n) +1, pos[i%n]
        elif words[i] not in lst:
            lst.add(words[i])
        else:
            return (i % n) +1, pos[i%n]
    return [0,0]

나는 position을 행렬을 만들어서 식을 넣었지만

몫을 계산해서 몇번째 행인지 알 수 있음

 

def solution(n, words):
    for p in range(1, len(words)):
        if words[p][0] != words[p-1][-1] or words[p] in words[:p]: return [(p%n)+1, (p//n)+1]
    else:
        return [0,0]

참고코드

https://school.programmers.co.kr/learn/courses/30/lessons/12981/solution_groups?language=python3