I had this programming problem that I'm trying to solve and since I'm running out of patience, instead of coding the script to follow an algorithm (which I wasnt able to code perfectly), I resorted to solve it using the pattern I see in the problem.

It works don't get me wrong, but I do wonder if it is generally alright to solve a problem (that I think is intended to train your brain to create a general algorithm from a series of patterns) with a solution that is derived from how the pattern works.

And the pattern I saw is that since the pattern on how the spiral is made is this:

N (1s to the right)
N-1 (1s downwards)
N-1 (1s to the right)
N-3 (1s upwards)
N-3 (1s to the left)
.. and so on and so forth.

Until it is the 2nd n-x where the difference is 2, or the last n-x where the difference is 1, I keep on printing 1s based on the pattern above to the direction that I'm also looping.

I solved it until the attempt and is able to submit but I felt that I cheated the problem.

I'm not asking for the best algorthm for the problem, but for your thoughts on how I did this problem. Was it a good alternative? As long is it works, it's an algorithm?

I don't know where else I can ask this question and this is a little broad but I hope I can get some ideas here. Thanks.

(Added JavaScript since I did my solution there and the one I have experience the most)

EDIT:

• Adding solution for a clearer understanding of the pattern I noticed.

function spiralize (n) {
  var output = Array(n).fill(0).map(_ => Array(n).fill(0));
  var i = 0, j = -1, max = n;
  while(true) {
    if([[0,1], [1,0], [0,-1], [-1, 0]].find(([x, y], index) => {
      [output, i, j, max] = traverseArray(output, i, j, max, x, y, n);
      if((index % 2 == 0 && max == 2) || (index % 2 == 1 && max == 1))
        return true;
    }))
      return output;
  }
}

function traverseArray(output, i, j, max, iter1, iter2, n) {
  iter1 ? max == n ? max-- : max -= 2 : max;
  temp = max;
  while(temp-- > 0)
    output[i += iter1][j += iter2] = 1;    
  return [output, i, j, max];
}