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g1201_1300.s1266_minimum_time_visiting_all_points.readme.md Maven / Gradle / Ivy

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:**

![](https://assets.leetcode.com/uploads/2019/11/14/1626_example_1.PNG)

**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`