[LeetCode]#1266. Minimum Time Visiting All Points

  • 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.
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
Input: points = [[3,2],[-2,2]]
Output: 5
  • points.length == n
  • 1 <= n <= 100
  • points[i].length == 2
  • -1000 <= points[i][0], points[i][1] <= 1000
class Solution:
def minTimeToVisitAllPoints(self, points):
size=len(points)
total_t=0
for i in range(size-1):
x=abs(points[i][0]-points[i+1][0])
y=abs(points[i][1]-points[i+1][1])
total_t+=max(x,y)
return total_t

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Fatboy Slim

Fatboy Slim

Interesting in any computer science.