869. Reordered Power of 2

https://leetcode.com/problems/reordered-power-of-2/description/

Description

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.

Example 1:
Input: 1
Output: true

Example 2:
Input: 10
Output: false

Example 3:
Input: 16
Output: true

Example 4:
Input: 24
Output: false

Example 5:
Input: 46
Output: true

Solution

class Solution(object):
    def reorderedPowerOf2(self, N):
        """
        :type N: int
        :rtype: bool
        """
        from collections import Counter
        record = {
        (('1',2),('2',2),('3',1),('4',1),('7',2),('8',1)),
        (('1',1),('2',1),('8',1)),
        (('4',1),),
        (('2',1),),
        (('2',2),('4',1),('5',1),('8',2)),
        (('1',1),('6',1)),
        (('2',1),('3',1),('6',1),('7',1),('8',1)),
        (('1',2),('2',1),('6',2),('7',3)),
        (('0',1),('4',1),('6',1),('9',1)),
        (('2',1),('5',1),('6',1)),
        (('0',1),('2',1),('4',1),('8',1)),
        (('4',1),('6',1)),
        (('0',1),('1',1),('2',1),('4',1)),
        (('1',1),('2',1),('8',1),('9',1)),
        (('0',1),('1',1),('4',1),('6',2),('7',1),('8',2)),
        (('0',1),('1',2),('2',1),('3',1),('7',1)),
        (('0',1),('1',1),('4',1),('5',1),('6',1),('7',1),('8',1)),
        (('3',1),('5',2),('6',2)),
        (('2',1),('3',1),('4',2),('5',2),('6',2),('8',1)),
        (('0',1),('1',1),('3',1),('4',3),('9',1)),
        (('1',1),),
        (('1',1),('2',1),('5',1)),
        (('2',1),('3',3),('4',2),('5',2)),
        (('1',1),('2',2),('4',2),('6',1)),
        (('8',1),),
        (('1',1),('3',1),('4',1),('6',1),('8',1)),
        (('0',1),('3',1),('6',1),('8',4)),
        (('0',1),('1',1),('2',2),('5',1),('7',1),('9',1)),
        (('2',1),('3',1)),
        (('0',1),('1',1),('2',1),('3',1),('5',1),('6',1),('7',1),('8',1),('9',1))
        }
        return tuple(sorted(list(Counter(str(N)).items()))) in record

其中，record 变量是通过下列代码产生的：

from collections import Counter
a = []
i = 0
while True:
    if 2**i<=10**9:
        a.append(str(2**i))
    else:
        break
    i += 1
record = set([tuple(sorted(list(Counter(i).items()))) for i in a])
print(record)