LeetCode: Reordered Power Of 2 Solution

1/**
2 * @param {number} N
3 * @return {boolean}
4 */
5var reorderedPowerOf2 = function (N) {
6  const getDigits = number => {
7    const digits = Array(10).fill(0)
8    while (number) {
9      digits[number % 10]++
10      number = Math.floor(number / 10)
11    }
12    return digits.join("")
13  }
14
15  // 0 <= N <= 1e9
16  // 0 <= 2^NUMBER_OF_DIGITS <= 1e9
17  // 0 <= NUMBER_OF_DIGITS <= 30
18  const MAX_NUMBER_OF_DIGITS = Math.ceil(Math.log2(1e9))
19
20  for (let exp = 0; exp <= MAX_NUMBER_OF_DIGITS; exp++) {
21    if (getDigits(N) === getDigits(Math.pow(2, exp))) {
22      return true
23    }
24  }
25  return false
26}