869. Reordered Power of 2 - LeetCode Solution

class Solution:
    def reorderedPowerOf2(self, n: int) -> bool:
        for i in range(32):
            dict1 = {}
            
            
            binary = str(2**i)
            if len(binary) != len(str(n)):
                continue
            for digit in binary:
                if digit in dict1:
                    dict1[digit]+=1
                else:
                    dict1[digit] = 1
            dict2 = {}
            for digit in str(n):
                if digit in dict2:
                    dict2[digit] +=1
                else:
                    dict2[digit] = 1
        
            if dict1 == dict2:
                return True
        return False