1266 Minimum Time Visiting All Points - Easy

Problem:

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

Problem Analysis:

• this is a geometry question for the L∞ norm or Chebyshev Distance.
• Time Complexity: O(n) where n is the number of points in the list
• Space Complexity: O(1)

Solutions:

class Solution:
    def minTimeToVisitAllPoints(self, points: List[List[int]]) -> int:
        res = 0

        for i in range(len(points)-1):
            x1,y1 = points[i]
            x2,y2 = points[i+1]
            res += max(abs(x2-x1), abs(y2-y1))
        return res

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