getArraysOfDigits (challenge)

/**
 * Create a function `getArraysOfDigits(subarrayLength, step)` to generate an array of arrays of digits.
 * Each subarray must be the same length (`subarrayLength`) and contain numbers increasing by 1,
 * unless that number exceeds 9, in which case it must wrap back around to (number - 9).
 * The first subarray starts at 1.
 * The first digit of each subsequent subarray must be equal to the sum of `step` and the first digit in the previous subarray.
 * There must be as many subarrays as needed so that the first digit in the last subarray equals 1 again.
 * Examples and solution below.
 * /

/*
getArraysOfDigits(3, 3)
[
  [1, 2, 3],
  [4, 5, 6],
  [7, 8, 9],
  [1, 2, 3]
]

getArraysOfDigits(1, 2)
[
  [1],
  [3],
  [5],
  [7],
  [9],
  [2],
  [4],
  [6],
  [8],
  [1]
]

getArraysOfDigits(5, 2)
[
  [1, 2, 3, 4, 5],
  [3, 4, 5, 6, 7],
  [5, 6, 7, 8, 9],
  [7, 8, 9, 1, 2],
  [9, 1, 2, 3, 4],
  [2, 3, 4, 5, 6],
  [4, 5, 6, 7, 8],
  [6, 7, 8, 9, 1],
  [8, 9, 1, 2, 3],
  [1, 2, 3, 4, 5]
]
*/

/**
 *
 * @param {number} subarrayLength The length of each nested array.
 * @param {number} step By how much the first element in the current subarray will increase in the next subarray.
 * @returns {number[][]} An array as described above.
 */
function getArraysOfDigits(subarrayLength, step) {
  const result = [];
  const toDigit = (num) => (num - 1) % 9 + 1;

  for (let i = 1; ; i = toDigit(i + step)) {
    const newSubarray = Array.from({ length: subarrayLength }, (_, j) => {
      return toDigit(i + j);
    });
    result.push(newSubarray);

    if (result.length > 1 && i === 1)
      break;
  }

  return result;
}