💡 Problem Formulation: The challenge is to determine if a given number can be transformed into a power of two by reordering its digits. For instance, given the input 4612, the desired output is True, since the digits can be reordered to form 2048, which is a power of 2.
Method 1: Brute Force with String Manipulation
This method involves generating all permutations of the digits of a number and checking each permutation to determine if it’s a power of 2. We use itertools’ permutations to generate possible numbers and check each against a precomputed set of power-of-two strings.
Here’s an example:
from itertools import permutations
def reordered_power_of2(N):
N = str(N)
possible_numbers = set([''.join(p) for p in permutations(N) if p[0] != '0'])
power_of_2 = set(str(1 << i) for i in range(31))
return any(num in power_of_2 for num in possible_numbers)
print(reordered_power_of2(4612))Output: True
This code compresses each permutation that could represent a number (excluding those with a leading zero) into a set and then checks each to ascertain if it’s a power of 2. It’s straightforward but could be inefficient for large numbers due to the growth of permutations.
Method 2: Sorting Digits and Comparing
A more efficient approach involves sorting the digits of the number and comparing this sorted string to a sorted string of each power of 2. It is fast because it avoids the need to check every permutation.
Here’s an example:
def reordered_power_of2(N):
N_sorted = sorted(str(N))
return any(N_sorted == sorted(str(1 << i)) for i in range(31))
print(reordered_power_of2(4612))Output: True
This snippet sorts the digits of the input number and compares them to the sorted digits of powers of two. This method is considerably faster than brute force, as it relies on the property that two numbers with the same digits will have the same sorted digit sequence.
Method 3: Counting Digits
A refinement of the sorting method is to use digit counts. Importantly, two numbers are permutations of each other if and only if they have the same digit counts. This method uses a counting sort approach to compare digit frequencies.
Here’s an example:
from collections import Counter
def reordered_power_of2(N):
N_count = Counter(str(N))
return any(N_count == Counter(str(1 << i)) for i in range(31))
print(reordered_power_of2(4612))Output: True
The Counter class from the collections module tallies the occurrences of each digit. We compare the digit count of the given number with the digit count of powers of two. It’s efficient as comparisons based on digit count are quicker than string comparisons.
Method 4: Mathematical Properties
By utilizing mathematical properties, we can narrow down the powers of 2 that need to be checked. For example, we only need to consider powers of 2 with the same number of digits as the input number.
Here’s an example:
def reordered_power_of2(N):
N_str = sorted(str(N))
length = len(N_str)
min_power = 10**(length - 1)
max_power = 10**length - 1
i = 0
while (1 << i) < min_power:
i += 1
while (1 << i) <= max_power:
if sorted(str(1 << i)) == N_str:
return True
i += 1
return False
print(reordered_power_of2(4612))Output: True
This method calculates the range of powers of 2 that need to be checked, based on the number of digits in the input. It sorts the input once and checks against the sorted powers of 2 within this range, reducing the computation required significantly.
Bonus One-Liner Method 5: Using Set and Map
For a minimalistic solution, the same principle can be expressed in a single line of Python code. The following achieves digit count comparison using a set and a map comprehension.
Here’s an example:
print(sorted(str(4612)) in map(lambda i: sorted(str(1 << i)), range(31)))
Output: True
This one-liner performs a sorted comparison of the digits of the number with those of all powers of 2 up to 2^30. It’s elegant and utilizes functional programming principles but may be slightly less readable for beginners.
Summary/Discussion
- Method 1: Brute Force with String Manipulation. This is conceptually simple but computationally expensive due to factorial complexity.
- Method 2: Sorting Digits and Comparing. Much more efficient than brute force; however, it does involve multiple sort operations that have a complexity of O(n log n).
- Method 3: Counting Digits. A further optimization of Method 2 where digit counting leads to an even faster comparison, with reduced time complexity.
- Method 4: Mathematical Properties. This method leverages number theory and optimizes the search space for candidate powers of 2, thus is efficient for large numbers.
- Method 5: Using Set and Map. This one-liner, while concise, may sacrifice a bit of clarity but impresses with its elegant use of Python’s functional aspects.