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🚀 이슈 번호
Resolve: {#2328}
🧩 문제 해결
스스로 해결: ✅
🔎 접근 과정
다른 문제들과 같이 똑같이 BFS를 사용하되,
회전에 따른 visited를 0, 1로 구분하여 처리하였다.
⏱️ 시간 복잡도
O(N^4)visited 배열로 인해 이미 방문한 곳은 새로 방문되지 않는다.
이로 인해 각 board를 한 번 씩만 반복하기 때문에 N^2이지만,
회전에 따른 방문을 또 진행하기 때문에 N^4이 된다.
💻 구현 코드