프로그래머스 - 코딩테스트 연습 - 스택/큐 - 프린터

문제

https://school.programmers.co.kr/learn/courses/30/lessons/42587

정답

공간 복잡도 낮은 방법

def solution(priorities, location):
    order = 1
    max_p = max(priorities)

    while priorities:
        priority = priorities.pop(0)

        if max_p == priority:
            if location == 0:
                return order
            max_p = max(priorities)
            order += 1
        else:
            priorities.append(priority)

        if location == 0:
            location = len(priorities) - 1
        else:
            location -= 1

간결한 방법

def solution(priorities, location):
    queue =  list(enumerate(priorities))
    order = 0
    while queue:
        cur = queue.pop(0)
        if any(cur[1] < q[1] for q in queue):
            queue.append(cur)
        else:
            order += 1
            if cur[0] == location:
                return order