869.reordered power of 2

Starting with a positive integer N, we reorder the digits in any order (including the original order) such that the leading digit is not zero.

Return true if and only if we can do this in a way such that the resulting number is a power of 2.

Input: 1
Output: true

Input: 10
Output: false

Input: 16
Output: true

Input: 24
Output: false

Input: 46
Output: true

• 1 <= N <= 10^9

1. updating

class Solution:
def reorderedPowerOf2(self, N):
    li=[]
    for i in range(30):
        li.append(str(2**i))
    def judge(a,b):
        if(len(a) == len(b)):
            A=list(a)
            B=list(b)
            for index in A:
                if(index in B):
                    B.remove(index)
                else:
                    return False
            return True
        return False
    for i in range(len(li)):
        if(judge(li[i],str(N))):
            return True
    return False