Introduction

In this blog post, we will explore how to solve the LeetCode problem "1503" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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. 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. 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 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. 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.

Explanation

Here's the solution to determine the moment when the last ant falls off the plank:

• Key Insight: When two ants meet and change directions, it's equivalent to them passing through each other. We only care about the time it takes for each ant to reach the end of the plank.
• Approach: Calculate the time each ant takes to fall off the plank, either by moving left or right. The maximum of these times will be the time when the last ant falls.

• Optimization: Iterate through the left and right arrays, compute the time for each ant, and keep track of the maximum time.

• Complexity: Time Complexity: O(m + k), where m is the length of left and k is the length of right. Space Complexity: O(1).

Code

    def lastMoment(n: int, left: list[int], right: list[int]) -> int:
    """
    Calculates the moment when the last ant falls off the plank.

    Args:
        n: The length of the plank.
        left: A list of the positions of ants moving to the left.
        right: A list of the positions of ants moving to the right.

    Returns:
        The moment when the last ant falls off the plank.
    """

    max_time = 0

    # Calculate the time for ants moving to the left
    for pos in left:
        max_time = max(max_time, pos)

    # Calculate the time for ants moving to the right
    for pos in right:
        max_time = max(max_time, n - pos)

    return max_time