Package g1201_1300.s1266_minimum_time_visiting_all_points

Class Solution

java.lang.Object

g1201_1300.s1266_minimum_time_visiting_all_points.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 Summary

Constructors

Constructor	Description
Solution()

Method Summary

Modifier and Type	Method	Description
int	minTimeToVisitAllPoints(int[][] points)

Methods inherited from class java.lang.Object

clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait

Constructor Details

Solution

public Solution()

Method Details

minTimeToVisitAllPoints

public int minTimeToVisitAllPoints(int[][] points)