# Minimum Time Visiting All Points

## Problem

Given a `2D`

plane (cartesian coordinate system), there are `n`

points with integer coordinates `points[i] = [xi, yi]`

. Our task is to find the minimum time in seconds to visit all points.

We can move according to the below rules:

- In
`1`

second, we can move either one unit vertically, one unit horizontally or diagonally (it means we move one unit vertically and one unit horizontally in one second). - We have to visit the points in the same order as they appear in the array.
- We are allowed to pass through points that appear later in the order, but these do not count as visits.

## Input

**Example 1:**

Input:points = [[1,1],[3,4],[-1,0]]Output:7Explanation: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

**Example 3:**

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`

## Solution

### First Solution

The optimal way to move from point `A`

to point `B`

is to move diagonally as a shortcut. The rest is to move in a straight line.

```
public int MinTimeToVisitAllPoints(int[][] points)
{
int min = 0, diffX = 0, diffY = 0;
for (int i = 0; i < points.Length - 1; i++)
{
diffX = Math.Abs(points[i + 1][0] - points[i][0]);
diffY = Math.Abs(points[i + 1][1] - points[i][1]);
min += (diffX < diffY) ? diffX : diffY;
min += Math.Abs(diffX - diffY);
}
return min;
}
```

### Second Solution

The second solution is slightly faster than the first solution because we use less arithmetic operation. The idea here is that when we add our `min`

with `Math.Abs(diffX - diffY)`

which is the rest straight line that we need to move, it is basically the same to finding the maximum different between `diffX`

and `diffY`

.

```
public int MinTimeToVisitAllPoints(int[][] points) {
int min = 0, diffX = 0, diffY = 0;
for (int i = 0; i < points.Length - 1; i++)
{
diffX = Math.Abs(points[i + 1][0] - points[i][0]);
diffY = Math.Abs(points[i + 1][1] - points[i][1]);
min += Math.Max(diffX, diffY);
}
return min;
}
```

## Time Complexity

The Time Complexity is * O(n)* because we have one loop where we iterate the input array which is

`points`

.## Space Complexity

The Time Complexity is * O(1)* because we don’t allocate memory with length

`n`

in our algorithm. We exclude array `points`

because this is the input.