Problem Description
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.
Key Insights
- Ants moving in opposite directions can be treated as if they pass through each other without changing the outcome.
- The time for an ant to fall off the plank is determined by its distance from the nearest end.
- For ants moving left, the time to fall is simply their position.
- For ants moving right, the time to fall is the distance from their position to the end of the plank (n).
Space and Time Complexity
Time Complexity: O(m + k), where m is the number of ants moving left and k is the number of ants moving right.
Space Complexity: O(1), as we are using a constant amount of space.
Solution
To solve this problem, we evaluate the maximum time it takes for any ant to fall off the plank. For ants moving left, the time to fall is equal to their position. For ants moving right, the time to fall is equal to the length of the plank minus their position. We simply calculate these values and return the maximum.