Class Solution


  • public class Solution
    extends Object
    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
    • Constructor Detail

      • Solution

        public Solution()
    • Method Detail

      • minTimeToVisitAllPoints

        public int minTimeToVisitAllPoints​(int[][] points)