1266. Minimum Time Visiting All Points

Problem

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&nbsp;<= 100

• points[i].length == 2

• -1000&nbsp;<= points[i][0], points[i][1]&nbsp;<= 1000

Solution (Java)

class Solution {
    public int minTimeToVisitAllPoints(int[][] points) {
        int minTime = 0;
        for (int i = 0; i < points.length - 1; i++) {
            minTime += chebyshevDistance(points[i], points[i + 1]);
        }
        return minTime;
    }

    private int chebyshevDistance(int[] pointA, int[] pointB) {
        return Math.max(Math.abs(pointA[0] - pointB[0]), Math.abs(pointA[1] - pointB[1]));
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).