Problem URL: Reordered power of 2

We will count the number of digits in out given input. Then we take all the power of 2 from 1 to 32 bit integers, compare the digits of them, if any one of it matches with our given input, we return true, otherwise false.

class Solution:
    def reorderedPowerOf2(self, n: int) -> bool:
        count = Counter(str(n))
        return any(count == Counter(str(1 << i)) for i in range(31))

Time Complexity: O(n^2)
Space Complexity: O(1)