LeetCode-in-Java

1266. Minimum Time Visiting All Points

Easy

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

Solution

public 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]));
    }
}

This site is open source. Improve this page.