The “Minimum Time Visiting All Points” problem on Leetcode is a fascinating problem that combines elements of geometry with algorithmic thinking. While the problem may seem deceptively simple at first glance, it offers opportunities for understanding deeper computational concepts. In this article, we’ll explore the problem in-depth, looking at various ways to approach its solution, discussing the computational complexity of these approaches, and providing Python code for each method.

## Problem Statement

### Description

You are given an array `points`

where `points[i] = [xi, yi]`

represents the coordinates of the ith point on a 2D plane. Multiple points can have the same coordinates. You are allowed to move any point at any time in one of the eight cardinal directions (left, right, up, down, or any of the four diagonals).

You need to return the minimum time in seconds to visit all the points in the order given by `points`

.

## Understanding the Problem

The problem essentially asks you to find the minimum time required to traverse from one point to the next in the given list of points on a 2D plane. You are allowed to move in any of the eight directions. This inherently implies that you can move diagonally, which can be a time-saving operation.

## Approaches

### Approach 1: Using Manhattan Distance with Diagonal Movement

A naive approach to this problem would be to calculate the minimum time required to traverse from one point to the next using the Manhattan Distance formula ∣x2−x1∣+∣y2−y1∣. However, because diagonal movement is allowed, we can improve this by traversing diagonally whenever possible.

#### Algorithm

- Initialize a variable
`total_time`

to zero. - For each pair of consecutive points (x1,y1) and (x2,y2):
- Calculate the minimum time required to go from (x1,y1) to (x2,y2) considering diagonal movement.
- Add this time to
`total_time`

.

#### Python Code Implementation

```
def min_time_to_visit_all_points(points):
total_time = 0
for i in range(len(points) - 1):
x1, y1 = points[i]
x2, y2 = points[i+1]
dx = abs(x2 - x1)
dy = abs(y2 - y1)
total_time += max(dx, dy)
return total_time
# Test the function
print(min_time_to_visit_all_points([[1,1],[3,4],[-1,0]])) # Output should be 7
```

#### Time Complexity

The time complexity of this algorithm is O(n), where n is the number of points.

#### Space Complexity

The space complexity is O(1).

### Approach 2: Utilizing Vector Operations (For Educational Purpose)

Although this problem doesn’t really benefit from this approach in terms of time or space complexity, it can be a good educational exercise to think about how vectors can be used to solve such problems.

#### Algorithm

- Initialize a variable
`total_time`

to zero. - For each pair of consecutive points A and B:
- Calculate the vector from A to B.
- Decompose this vector into its constituent diagonal vectors.
- Add the time required to traverse these vectors to
`total_time`

.

#### Python Code Implementation

```
import numpy as np
def min_time_to_visit_all_points(points):
total_time = 0
for i in range(len(points) - 1):
A = np.array(points[i])
B = np.array(points[i+1])
vec = B - A
total_time += np.linalg.norm(vec, ord=np.inf)
return int(total_time)
# Test the function
print(min_time_to_visit_all_points([[1,1],[3,4],[-1,0]])) # Output should be 7
```

#### Time Complexity

The time complexity remains O(n).

#### Space Complexity

The space complexity is O(1).

## Conclusion

The “Minimum Time Visiting All Points” problem on Leetcode is an interesting problem that combines elements of geometry, vector algebra, and basic algorithmic skills. While it’s straightforward enough to be solved by a simple loop and mathematical operations, it serves as a good introduction to the kinds of problems that can be tackled through a variety of mathematical tools.