1266. Minimum Time Visiting All Points

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

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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