백준 온라인 저지, 그리디 / 11497번: 통나무건너뛰기 (파이썬)

문제

남규는 통나무를 세워 놓고 건너뛰기를 좋아한다. 그래서 N개의 통나무를 원형으로 세워 놓고 뛰어놀려고 한다. 남규는 원형으로 인접한 옆 통나무로 건너뛰는데, 이때 각 인접한 통나무의 높이 차가 최소가 되게 하려 한다.

통나무 건너뛰기의 난이도는 인접한 두 통나무 간의 높이의 차의 최댓값으로 결정된다. 높이가 {2, 4, 5, 7, 9}인 통나무들을 세우려 한다고 가정하자. 이를 [2, 9, 7, 4, 5]의 순서로 세웠다면, 가장 첫 통나무와 가장 마지막 통나무 역시 인접해 있다. 즉, 높이가 2인 것과 높이가 5인 것도 서로 인접해 있다. 배열 [2, 9, 7, 4, 5]의 난이도는 |2-9| = 7이다. 우리는 더 나은 배열 [2, 5, 9, 7, 4]를 만들 수 있으며 이 배열의 난이도는 |5-9| = 4이다. 이 배열보다 난이도가 낮은 배열은 만들 수 없으므로 이 배열이 남규가 찾는 답이 된다.

입력

입력은 T개의 테스트 케이스로 이루어져 있다. 첫 줄에 T가 주어진다.

이어지는 각 줄마다 첫 줄에 통나무의 개수를 나타내는 정수 N(5 ≤ N ≤ 10,000), 둘째 줄에 각 통나무의 높이를 나타내는 정수 L_i가 주어진다. (1 ≤ L_i ≤ 100,000)

출력

각 테스트 케이스마다 한 줄에 주어진 통나무들로 만들 수 있는 최소 난이도를 출력하시오.

제한

 

예제 입력 1

3 7 13 10 12 11 10 11 12 5 2 4 5 7 9 8 6 6 6 6 6 6 6 6 

예제 출력 1

1 4 0 

힌트

 

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

접근 방법

- 0. 통나무의 길이를 오름차순으로 정렬한다.
- 1. 정렬된 통나무를 하나씩 탐색하며 왼쪽으로 추가할 나무와 오른쪽으로 추가할 나무를 하나씩 고려하여 현재 탐색 중인 통나무와 높이 차이가 더 많이 나는 쪽(숫자가 더 작은 쪽)에 해당 통나무를 설치한다.
- 2. 위의 과정을 반복한 뒤, 모든 탐색을 마치면 최대 높이를 출력한다.

코드

# https://www.acmicpc.net/problem/11497
# 접근 방법 (실버1)
# 0. 통나무의 길이를 오름차순으로 정렬한다.
# 1. 정렬된 통나무를 하나씩 탐색하며 왼쪽으로 추가할 나무와 오른쪽으로 추가할 나무를 하나씩 고려하여 현재 탐색 중인 통나무와 높이 차이가 더 많이 나는 쪽(숫자가 더 작은 쪽)에 해당 통나무를 설치한다. 
# 2. 위의 과정을 반복한 뒤, 모든 탐색을 마치면 최대 높이를 출력한다.

tc = int(input())
for _ in range(tc):
    n = int(input())
    l = list(map(int, input().split()))
    l.sort()
    left, right = l[0], l[0]
    min_difficulty = 0
    for i in range(1, n):
        if left <= right:
            min_difficulty = max(min_difficulty, l[i] - left)
            left = l[i]
        else:
            min_difficulty = max(min_difficulty, l[i] - right)
            right = l[i]

    print(min_difficulty)