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# 1266. Minimum Time Visiting All Points

<https://leetcode.com/problems/minimum-time-visiting-all-points>

## Description

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`

## ac

```java
```
