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

Note:

  1. 1 <= N <= 10^9

Solutions

1. Hash Table

# Time: O(n)
# Space: O(1)
class Solution:
    def reorderedPowerOf2(self, N: int) -> bool:
        key = self.counter(N)
        for i in range(32):
            if self.counter(1 << i) == key:
                return True
        return False
    
    def counter(self, num):
        res = 0
        while num != 0:
            res += pow(10, num % 10)
            num //= 10
        return res
# 135/135 cases passed (32 ms)
# Your runtime beats 56.63 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (12.8 MB)

References

  1. 869. Reordered Power of 2