1266. 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:

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

Solution: Math

• Time complexity: O(n)
• Space complexity: O(1)

 

JavaScript

/**
 * @param {number[][]} points
 * @return {number}
 */
const minTimeToVisitAllPoints = function (points) {
  const n = points.length;
  let result = 0;

  for (let index = 1; index < n; index++) {
    const [x1, y1] = points[index - 1];
    const [x2, y2] = points[index];
    const diffX = Math.abs(x1 - x2);
    const diffY = Math.abs(y1 - y2);

    result += Math.max(diffX, diffY);
  }

  return result;
};