Alpha 27 JS, Loops, Homework Problem 14

Hello everyone.
I have been struggling with this one for a few days now.
Here is the description:

Problem 14: Spiral Matrix

Write a program that reads from the console a positive integer number n (1 ≤ n ≤ 20) and prints a matrix holding the numbers from 1 to n*n in the form of square spiral like in the examples below.

n matrix
2 1 2
4 3
n matrix
3 1 2 3
8 9 4
7 6 5

Here is my 1/2 solution. It works from 1-10, but any larger number gives issues.

I have determined that the problem is some sort of Array limit of 100.
As you can see on the last line I have tried to overcome it with:
console.dir(arrA, {‘maxArrayLength’: null});
But still it doesn’t do the job.

I know there are other solutions, but I want to make this code work. Please, give me a way to deal with this JS limit.

My code is not explained in detail, so I will explain what happens in general.

  1. I create all the needed numbers from 1 to n*n but backwards.
  2. I create an array grid.
  3. Then I create a pattern based on the direction and number of indexes the spiral will ‘move’ from the center to the start.
    3.1 - create the letters corresponding to the directions, r, u, l, d.
    3.2 - create the corresponding numbers 1, 1, 2, 2…
    3.3 - combine the 2 arrays in r1, u1, l2, d2…
  4. Fill the grid with the corresponding numbers.
    There are different patterns for even and odd numbers, but that is covered. There are different starting centers for even and odd numbers which are also covered.

The code get’s each number from “numbers” and the direction from “finalPattern”, then according the the 4 If cases, it uses them to fill the grid.

  1. Converts the array in to a string, then splits it in to the appropriate number of rows.

Also, do you have any, free code sharing platfrom that doesn’t expire in 24 hours?

Thank you for your time!


The problem lies in the following line where you take out the steps into a variable when filling the matrix.
Let’s take for example, a finalPattern built for the input 11:
Taking the part of the string on position index 1 from finalPattern[i] works fine when the input is <= 10, because you won’t have two-digit numbers in your final pattern. The steps from end to end will always be less than 10. If your input is bigger than 10, you will have two-digit numbers in your finalPattern, which causes the bug - for example from r10, you take out 1, not 10 and that’s where your matrix stops filling. A suggested fix can be:
It’s not really scalable, as it won’t work for for three-digit numbers(you will have finalPatterns like r100 etc, but it was something I came up with in the moment. Other way you can fix it and make it work for any input is using an regex to extract the numbers from the string and initializing them into the steps variable. Good luck.

1 Like

I have made the change to the line that reads the “steps” and it works fine now.
Now it is with slice as you have suggested, but the end value is ‘finalPattern.length-1’. That way regardless of the number size it will still work.
Thank you a lot for your help!
If not for that I would still be searching in the wrong place for a problem that does not exist. :slight_smile:

1 Like