# 기둥과 보

> <https://programmers.co.kr/learn/courses/30/lessons/60061>

* 기둥이든 보든, 설치되고 삭제되는 기능만 있기 때문에, 2차원배열상에 표시할 필요없다.
  * 기둥과, 보 정보들을 가진 1차원 리스트를 활용하여 각 정보들의 존재여부 검증은 v in list 로 해주면 된다.
* 차례대로 삭제나 설치 일단 해보고
  * 해당 작업이 불가능 하다면 무르기

```python
def isPossible(r):
    for y, x, w in r:
        # 보라면
        if w:
            if [y, x - 1, 0] not in r and [y + 1, x - 1, 0] not in r and not ([y - 1, x, 1] in r and [y + 1, x, 1] in r):
                return True
        # 기둥이라면
        else:
            if x != 0 and [y, x, 1] not in r and [y - 1, x, 1] not in r and [y, x - 1, 0] not in r:
                return True

    return False


def solution(n, build_frame):

    r = []

    i = 1

    for y, x, w, t in build_frame:
        i += 1
        s = [y, x, w]
        # 설치
        if t:
            r.append(s)
            if isPossible(r):
                r.remove(s)

        # 삭제이면
        elif t == 0 and s in r:
            r.remove(s)
            if isPossible(r):
                r.append(s)

    r = sorted(r, key=lambda x: (x[0], x[1], x[2]))

    return r
```


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