Introduction

In this blog post, we will explore how to solve the LeetCode problem "1266" 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

On a 2D plane, there are n points with integer coordinates points[i] = [xi, yi]. Return the minimum time in seconds to visit all the points in the order given by points. You can move according to these rules: In 1 second, you can either: move vertically by one unit, move horizontally by one unit, or move diagonally sqrt(2) units (in other words, move one unit vertically then one unit horizontally in 1 second). You have to visit the points in the same order as they appear in the array. You are allowed to pass through points that appear later in the order, but these do not count as visits. Example 1: Input: points = [[1,1],[3,4],[-1,0]] Output: 7 Explanation: One optimal path is [1,1] -> [2,2] -> [3,3] -> [3,4] -> [2,3] -> [1,2] -> [0,1] -> [-1,0] Time from [1,1] to [3,4] = 3 seconds Time from [3,4] to [-1,0] = 4 seconds Total time = 7 seconds Example 2: Input: points = [[3,2],[-2,2]] Output: 5 Constraints: points.length == n 1 <= n <= 100 points[i].length == 2 -1000 <= points[i][0], points[i][1] <= 1000

Explanation

Here's the breakdown of the solution:

• Optimal Path: The fastest way to travel between two points is to move diagonally as much as possible and then move either horizontally or vertically to cover the remaining distance.
• Time Calculation: The time to move from point A to point B is determined by the maximum of the absolute differences in their x and y coordinates.

• Accumulate Time: Iterate through the points and accumulate the time taken to travel from the previous point to the current point.

• Runtime Complexity: O(n), where n is the number of points. Storage Complexity: O(1).

Code

    def minTimeToVisitAllPoints(points):
    """
    Calculates the minimum time to visit all points in the given order.

    Args:
        points (list of list of int): A list of points, where each point is a list [x, y].

    Returns:
        int: The minimum time to visit all points.
    """
    total_time = 0
    for i in range(1, len(points)):
        prev_x, prev_y = points[i - 1]
        curr_x, curr_y = points[i]
        total_time += max(abs(curr_x - prev_x), abs(curr_y - prev_y))
    return total_time