A non-cyclical number is an integer defined by the following algorithm:

• Given a positive integer, replace it with the sum of the squares of its digits.
• Repeat the above step until the number equals 1, or it loops infinitely in a cycle which does not include 1.
• If it stops at 1, then the number is a non-cyclical number.

Given a positive integer n, return true if it is a non-cyclical number, otherwise return false.

Example 1:

Input: n = 100

Output: true

Explanation: 1^2 + 0^2 + 0^2 = 1

Example 2:

Input: n = 101

Output: false

Explanation: 1^2 + 0^2 + 1^2 = 2 2^2 = 4 4^2 = 16 1^2 + 6^2 = 37 3^2 + 7^2 = 58 5^2 + 8^2 = 89 8^2 + 9^2 = 145 1^2 + 4^2 + 5^2 = 42 4^2 + 2^2 = 20 2^2 + 0^2 = 4 (This number has already been seen)

Constraints:

• 1 <= n <= 1000

Recommended Time & Space Complexity

You should aim for a solution as good or better than O(logn) time and O(logn) space, where n is the given integer.

Hint 1

Create a helper function that returns the sum of the squares of a number's digits. Then, simulate the given process. If we reach 1, return true. However, we may get stuck in a cycle if a number is processed more than once. What data structure can be used to detect if a number has already been processed?

Hint 2

We can use a hash set to detect if a number has already been processed. At each step, we update n with the return value of the helper function. If the result is 1, we return true. If n is already in the set, we return false. Otherwise, we add n to the hash set and continue.