> For the complete documentation index, see [llms.txt](https://mayanktyagi3111.gitbook.io/interview-prep/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://mayanktyagi3111.gitbook.io/interview-prep/strings-arrays-and-2-pointers/min-steps-in-infinite-grid.md).

# Min Steps in Infinite Grid

You are in an infinite 2D grid where you can move in any of the 8 directions

```
 (x,y) to 
    (x+1, y), 
    (x - 1, y), 
    (x, y+1), 
    (x, y-1), 
    (x-1, y-1), 
    (x+1,y+1), 
    (x-1,y+1), 
    (x+1,y-1) 
```

You are given a sequence of points and **the order in which you need to cover the points.**. Give the minimum number of steps in which you can achieve it. You start from the first point.

**NOTE:** This question is intentionally left slightly vague. Clarify the question by trying out a few cases in the “See Expected Output” section.

\
\
**Input Format**<br>

Given two integer arrays A and B, where A\[i] is x coordinate and B\[i] is y coordinate of ith point respectively.\
\
**Output Format**<br>

Return an Integer, i.e minimum number of steps.\
\
**Example Input**<br>

Input 1:

```
 A = [0, 1, 1]
 B = [0, 1, 2]
```

\
**Example Output**\
Output 1:

```
 2
```

**Example Explanation**\
Explanation 1:

```
 Given three points are: (0, 0), (1, 1) and (1, 2).
 It takes 1 step to move from (0, 0) to (1, 1). It takes one more step to move from (1, 1) to (1, 2).
```

```java
public class Solution {
    public int coverPoints(ArrayList<Integer> A, ArrayList<Integer> B) {
        int steps=0;
        for(int i=0;i<A.size()-1;i++){
            int diffx=A.get(i+1)-A.get(i);
            int diffy=B.get(i+1)-B.get(i);
            if(diffx==0 || diffy==0){
                steps+=Math.abs(diffx)+Math.abs(diffy);
            }
            else if(Math.abs(diffx)==Math.abs(diffy)){
                steps+=Math.abs(diffx);
            }
            else{
                steps+=Math.min(Math.abs(diffx),Math.abs(diffy))+Math.abs(Math.abs(diffx)-Math.abs(diffy));
            }
        }
        return steps;
    }
}

```


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