#1503

Medium

LeetCode Problem Workspace

Last Moment Before All Ants Fall Out of a Plank

This problem involves simulating ant movement on a plank to determine the last moment before all ants fall off.

Problem Statement

You are given a wooden plank of length n units, and some ants are walking on the plank. Each ant moves at a speed of 1 unit per second, and some ants move towards the left, while others move towards the right. The positions of the ants moving left are given in the array left, and those moving right are in the array right.

When two ants meet while walking in opposite directions, they simply swap directions without any time loss. Your task is to calculate the last moment before all ants fall off the plank. That is, determine the time when the last ant exits the plank.

Examples

Example 1

Input: n = 4, left = [4,3], right = [0,1]

Output: 4

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).

Example 2

Input: n = 7, left = [], right = [0,1,2,3,4,5,6,7]

Output: 7

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

All ants are going to the left, the ant at index 7 needs 7 seconds to fall.

Constraints

1 <= n <= 104
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.

Solution Approach

Find Maximum Time for Each Direction

The key observation here is that the ants don't actually need to swap directions when they meet. We can simulate the movement by considering that the ants pass through each other. The problem boils down to finding the maximum time for any ant to reach the end of the plank, either from the left or the right.

Use Maximum of Left and Right Arrays

For each ant, calculate the time it will take to fall off the plank. For ants in the left array, the time is simply their position. For ants in the right array, the time is the distance from the plank's right end (i.e., n - position). The last moment will be the maximum time of these values.

Return the Maximum Time

The final solution is the maximum of the times for ants in the left and right arrays. We compute the maximum time for the left-moving ants and the right-moving ants, then return the greater of the two.

Complexity Analysis

Metric	Value
Time	O(n + m)
Space	O(1)

The time complexity of this solution is O(n + m), where n is the length of the plank, and m is the number of ants (sum of the lengths of left and right). We iterate over each ant once to compute the maximum times, and the space complexity is O(1) since we only need a few variables to store the maximum times and input arrays.

What Interviewers Usually Probe

The candidate demonstrates an understanding of array manipulation and problem-solving techniques.
The candidate can identify key optimizations like ignoring the ant direction swap.
The candidate efficiently determines the maximum time by focusing on edge cases.

Common Pitfalls or Variants

Common pitfalls

Mistaking the direction swap as needing a complex simulation instead of a simple time calculation.
Not handling the case where there are no ants in either the left or right arrays.
Misunderstanding the problem as a simulation of movement, rather than focusing on maximum times.

Follow-up variants

The problem can be extended to consider more complicated interactions, such as different speeds for the ants.
A variation could involve multiple planks with ants walking between them, requiring a more complex simulation.
The problem can be modified to track not just the last moment, but also the number of ants left at different moments.

FAQ

How do I solve the Last Moment Before All Ants Fall Out of a Plank problem?

The solution involves calculating the maximum time each ant needs to reach the edge of the plank, ignoring direction changes upon meeting. The answer is the maximum time for any ant.

What is the time complexity of the Last Moment Before All Ants Fall Out of a Plank problem?

The time complexity is O(n + m), where n is the length of the plank and m is the total number of ants. The solution iterates over the left and right arrays once.

What is the key observation in the Last Moment Before All Ants Fall Out of a Plank problem?

The key observation is that when ants meet, they simply pass through each other, so we can treat their movements as independent without the need for direction changes.

What are the common pitfalls in solving the Last Moment Before All Ants Fall Out of a Plank problem?

Common pitfalls include overcomplicating the problem by simulating the ants' movements instead of focusing on their times to fall off the plank, and neglecting edge cases like no ants in one direction.

How does GhostInterview help with solving problems like Last Moment Before All Ants Fall Out of a Plank?

GhostInterview helps by focusing on optimizing the solution, recognizing patterns, and guiding through edge cases, ensuring candidates avoid common pitfalls.

terminal

Solution

Solution 1: Brain Teaser

The key point of the problem is that when two ants meet and then turn around, it is equivalent to the two ants continuing to move in their original directions. Therefore, we only need to find the maximum distance moved by any ant.

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class Solution:
    def getLastMoment(self, n: int, left: List[int], right: List[int]) -> int:
        ans = 0
        for x in left:
            ans = max(ans, x)
        for x in right:
            ans = max(ans, n - x)
        return ans

Continue Practicing

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