Overview

It seems that coding tests are conducted overseas in interviews with engineers, and in many cases, the main thing is to implement specific functions and classes according to the theme.

As a countermeasure, it seems that a site called Let Code will take measures.

A site that trains algorithmic power that can withstand coding tests that are often done in the home.

I think it's better to have the algorithm power of a human being, so I'll solve the problem irregularly and write down the method I thought at that time as a memo.

Leetcode

Basically, I would like to solve the easy acceptance in descending order.

Last time Leet Code Day4 starting from zero "938. Range Sum of BST"

problem

1266. Minimum Time Visiting All Points

Since multiple x-axis and y-axis coordinates are given in the list, the problem is to find the shortest time to visit all the coordinates.

As a rule, You can move one unit vertically, horizontally, or diagonally per second. The coordinates must be reached in the same order as given There is a restriction.

Looking at Example 1, you may not know how to find the shortest path in the graph, but as an important idea to solve this problem for the time being, find the shortest path. The point is that you can simply subtract from the numerical value with the larger difference from each given coordinate. The important thing here is not to get the difference as it is, but to get the absolute value.

Python has a built-in function, ʻabs () `, so you can just use it.

class Solution:
    def minTimeToVisitAllPoints(self, points: List[List[int]]) -> int:
        sec = 0
        x1,y1 = points.pop()
        while points:
            x2,y2 = points.pop()
            sec += max(abs(x2-x1),abs(y2-y1))
            x1,y1 = x2,y2
        return sec
# Runtime: 56 ms, faster than 84.71% of Python3 online submissions for Minimum Time Visiting All Points.
# Memory Usage: 13.9 MB, less than 100.00% of Python3 online submissions for Minimum Time Visiting All Points.

First of all, sec is an abbreviation for second, and prepares a variable for adding travel time.

And this time I used the pop of the stack. Python pop removes the trailing (last) element if no arguments are specified.

In the test case of the answer, [[1,1], [3,4], [-1,0]] is used, so in this case, -1,0 is substituted for x1, y1. 3 and 4 are assigned to x2 and y2, respectively. Then, after subtracting x1 from x2 and y1 from y2, use the max function to extract the larger number and assign it to sec. After that, x2 and y2 are assigned to x1 and y1, loop until the number of elements of points is exhausted, and finally sec, which is the sum of the maximum values of the difference, is returned.

I think the answer is simple as above.

You can also make an answer with a for statement.

class Solution:
    def minTimeToVisitAllPoints(self, points: List[List[int]]) -> int:
        sec = 0
        current = points[0]
        for point in points[1:]:
            sec += max(abs(current[0] - point[0]), abs(current[1] - point[1]))
            current = point
        return sec
# Runtime: 64 ms, faster than 32.15% of Python3 online submissions for Minimum Time Visiting All Points.
# Memory Usage: 13.9 MB, less than 100.00% of Python3 online submissions for Minimum Time Visiting All Points.

However, this is a little slow.

It would be nice to study the former if you also study the stack.

Let Code Day5 starting from zero "1266. Minimum Time Visiting All Points"

Overview

problem