README.md

1266. Minimum Time Visiting All Points

On a plane there are n points with integer coordinates points[i] = [xi, yi]. Your task is to find the minimum time in seconds to visit all points.

You can move according to the next rules:

• In one second always you can either move vertically, horizontally by one unit or diagonally (it means to move one unit vertically and one unit horizontally in one second).
• You have to visit the points in the same order as they appear in the array.

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

Solutions (Ruby)

1. Solution

# @param {Integer[][]} points
# @return {Integer}
def min_time_to_visit_all_points(points)
    ret = 0

    for i in 0...(points.length - 1)
        x0, y0 = points[i][0], points[i][1]
        x1, y1 = points[i + 1][0], points[i + 1][1]
        ret += [(x0 - x1).abs, (y0 - y1).abs].max
    end

    return ret
end

Solutions (Rust)

1. Solution

impl Solution {
    pub fn min_time_to_visit_all_points(points: Vec<Vec<i32>>) -> i32 {
        let mut ret = 0;

        for i in 0..(points.len() - 1) {
            let (x0, y0) = (points[i][0], points[i][1]);
            let (x1, y1) = (points[i + 1][0], points[i + 1][1]);
            ret += (x0 - x1).abs().max((y0 - y1).abs());
        }

        ret
    }
}