## 1266. Minimum Time Visiting All Points

On a 2D plane, there are `n`

points with integer coordinates `points[i] = [x`

. Return _{i}, y_{i}]*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:7Explanation: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`

## Rust Solution

```
struct Solution;
impl Solution {
fn min_time_to_visit_all_points(points: Vec<Vec<i32>>) -> i32 {
let mut res = 0;
let n = points.len();
for i in 1..n {
let x1 = points[i - 1][0];
let y1 = points[i - 1][1];
let x2 = points[i][0];
let y2 = points[i][1];
res += i32::max((x2 - x1).abs(), (y2 - y1).abs());
}
res
}
}
#[test]
fn test() {
let points = vec_vec_i32![[1, 1], [3, 4], [-1, 0]];
let res = 7;
assert_eq!(Solution::min_time_to_visit_all_points(points), res);
}
```

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