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🧷 문제 링크
https://www.acmicpc.net/problem/14719
🧭 풀이 시간
40분
👀 체감 난이도
✏️ 문제 설명
2차원 세계에서 쌓여있는 벽들 사이에 생길 수 있는 빗물의 개수를 출력
🔍 풀이 방법
구현
단순히 매 좌표마다 좌우를 모두 확인하여 빗물 개수를 세고 모든 좌표를 확인했을 때 카운트되는 빗물이 없다면 반복문 탈출하도록 구현
⏳ 회고
알고리즘을 배울수록 n^2은 나쁜거야! 이걸로 풀리게 내겠어? 라는 생각때문에 구현을 주저하는 경우가 많은 것 같다. 입력을 잘 살피고 가능하면 무지성으로 구현하는게 좋은듯..