1266.MinimumTimeVisitingAllPoints.py

'''
    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:
    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:
    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
'''
#Difficulty: Easy
#122 / 122 test cases passed.
#Runtime: 72 ms
#Memory Usage: 14.2 MB

#Runtime: 72 ms, faster than 13.93% of Python3 online submissions for Minimum Time Visiting All Points.
#Memory Usage: 14.2 MB, less than 61.36% of Python3 online submissions for Minimum Time Visiting All Points.

class Solution:
    def minTimeToVisitAllPoints(self, points: List[List[int]]) -> int:
        time = 0
        for i in range(len(points)-1):
            x = abs(points[i][0] - points[i+1][0])
            y = abs(points[i][1] - points[i+1][1])
            time += max(x, y)
        return time