[Leetcode] Last Moment Before All Ants Fall Out of a Plank (#1503)

 

 

문제 

 

We have a wooden plank of the length n units. Some ants are walking on the plank, each ant moves with a speed of 1 unit per second. Some of the ants move to the left, the other move to the right.

When two ants moving in two different directions meet at some point, they change their directions and continue moving again. Assume changing directions does not take any additional time.

When an ant reaches one end of the plank at a time t, it falls out of the plank immediately.

Given an integer n and two integer arrays left and right, the positions of the ants moving to the left and the right, return the moment when the last ant(s) fall out of the plank.

Input: n = 4, left = [4,3], right = [0,1]
Output: 4
Explanation: In the image above:
-The ant at index 0 is named A and going to the right.
-The ant at index 1 is named B and going to the right.
-The ant at index 3 is named C and going to the left.
-The ant at index 4 is named D and going to the left.
The last moment when an ant was on the plank is t = 4 seconds. After that, it falls immediately out of the plank. (i.e., We can say that at t = 4.0000000001, there are no ants on the plank).

 

해설 코드 

class Solution(object):
    def getLastMoment(self, n, left, right):
        
        if not right:
            return max(left)
        if not left:
            return n - min(right)
        
        l = max(left)
        r = n - min(right)
        
        return max(l,r)

분석 > 풀고나니 너무 쉬운 상식적인 문제라 허무했다. 왼쪽에서 오는 개미와 오른쪽에서 오는 개미가 겹치는 건 애초에 고려하지 않아도 되고. 부딪히나 마나 개미는 원래 자기가 가려던 방향으로 가는 것이기 때문에 가장 왼쪽에 있는 개미가 가장 오른쪽에 도착하는 시간과 가장 오른쪽에 있는 개미가 가장 왼쪽에 도착하는 시간 중에 더 큰 값을 반환하면, 그 값이 마지막 개미가 떨어지는 시간이었다. 

 

 

Last Moment Before All Ants Fall Out of a Plank - LeetCode