Spiral Numbers

Construct a square matrix with a size N × N containing integers from 1 to N * N in a spiral order, starting from top-left and in clockwise direction.

Example

For n = 3, the output should be

spiral_numbers(n) = [[1, 2, 3],
                    [8, 9, 4],
                    [7, 6, 5]]


 1  2  3  4
12 13 14  5
11 16 15  6
10  9  8  7


 1  2  3  4  5
16 17 18 19  6
15 24 25 20  7
14 23 22 21  8
13 12 11 10  9

Input/Output

• [input] integer n

  Matrix size, a positive integer.

Solution

PYTHON

def spiral_numbers(n):
    matrix = [[0] * n for _ in range(n)]
    h_row, h_col, v_row, v_col = 0, 0, 0, 1

    for matrix[h_row][h_col] in range(1, n * n + 1):
        if matrix[(h_row + v_row) % n][(h_col + v_col) % n]:
            v_row, v_col = v_col, -v_row
        h_row, h_col = (h_row + v_row), (h_col + v_col)
    return matrix


print(spiral_numbers(3))

JAVASCRIPT

function spiralNumbers(n) {
  let result = [];
  for (let i = 0; i < n; i++) {
    result.push([]);
  }
  let [x, y, dx, dy] = [0, 0, 1, 0];
  for (let i = 1; i <= n * n; i++) {
    result[y][x] = i;
    if (x + dx >= n || x + dx < 0 || y + dy >= n || y + dy < 0 || result[y + dy][x + dx] !== undefined) {
      let temp = dx;
      dx = -dy;
      dy = temp;
    }
    x += dx;
    y += dy;
  }
  return result;
}

console.log(spiralNumbers(3));