1503. Last Moment Before All Ants Fall Out of a Plank

We have a wooden plank of the length n units. Some ants are walking on the plank, each ant moves with speed 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 doesn't take any additional time.

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

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.

Example 1:

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.
Note that the last moment when an ant was on the plank is t = 4 second, after that it falls imediately out of the plank. (i.e. We can say that at t = 4.0000000001, there is no ants on the plank).

Example 2:

Input: n = 7, left = [], right = [0,1,2,3,4,5,6,7]
Output: 7
Explanation: All ants are going to the right, the ant at index 0 needs 7 seconds to fall.

Example 3:

Input: n = 7, left = [0,1,2,3,4,5,6,7], right = []
Output: 7
Explanation: All ants are going to the left, the ant at index 7 needs 7 seconds to fall.

Example 4:

Input: n = 9, left = [5], right = [4]
Output: 5
Explanation: At t = 1 second, both ants will be at the same intial position but with different direction.

Example 5:

Input: n = 6, left = [6], right = [0]
Output: 6

Constraints:

• 1 <= n <= 10^4

• 0 <= left.length <= n + 1

• 0 <= left[i] <= n

• 0 <= right.length <= n + 1

• 0 <= right[i] <= n

• 1 <= left.length + right.length <= n + 1

• All values of left and right are unique, and each value can appear only in one of the two arrays.

范围为[0, n]的一维坐标系，left表示对应位置的蚂蚁会向左走，right为向右走的，当两个蚂蚁相遇后会分别折返，问最后一只蚂蚁走出坐标系需要的步数。蚂蚁相遇后的效果依然是一只蚂蚁往左另一只往右，区分哪只是哪只并无裨益，因此分别统计往左走的离0有多远，右边离n多远，找它们中距离最大的

class Solution:
    def getLastMoment(self, n: int, left: List[int], right: List[int]) -> int:
        return max(max(left or [0]), n - min(right or [n]))