Minimum Time Visiting All Points

On a 2D plane, there are n points with integer coordinates points[i] = [x<sub>i</sub>, y<sub>i</sub>]. 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.

LeetCode Problem - 1266

class Solution {
    public int minTimeToVisitAllPoints(int[][] points) {
        int answer = 0; // Initialize the total time to visit all points

        // Iterate through the points array
        for (int i = 0; i < points.length - 1; i++) {
            int x1 = points[i][0]; // Get the x-coordinate of the current point
            int y1 = points[i][1]; // Get the y-coordinate of the current point

            int x2 = points[i + 1][0]; // Get the x-coordinate of the next point
            int y2 = points[i + 1][1]; // Get the y-coordinate of the next point

            // Calculate the time to move from the current point to the next point
            // Time is determined by the maximum of the absolute differences in x and y coordinates
            answer += Math.max(Math.abs(x2 - x1), Math.abs(y2 - y1));
        }

        return answer; // Return the total time to visit all points
    }
}