Problem URL: Encode number

The following sequence can be built up form the ealier result. So we can search index of the prefix part, for example:

f(5) = "10", f(6) = "11"

The prefix are both f(2) = "1"

so we found that f(n) has f((n - 1) / 2) as prefix.

class Solution:
    def encode(self, num: int) -> str:
        return self.encode((num-1)//2) + '10'[num % 2] if num else ""

Time complexity: O(log(n))
Space complexity: O(log(n))