**π‘ Problem Formulation:** You may encounter a situation where you need to determine if the binary representation of two numbers are anagrams of each other. Essentially, an anagram in this context means the binary strings have the same frequency of ‘1’s and ‘0’s. For example, if you take the numbers 5 (binary 101) and 6 (binary 110), they are anagrams in binary form since they both contain two ‘1’s and one ‘0’.

## Method 1: Sort and Compare

This method involves converting the numbers to binary, stripping the ‘0b’ prefix, sorting the characters, and finally comparing the sorted strings. While not the most efficient, it is straightforward and easy to understand. If the sorted strings are equal, the binary representations are anagrams.

Here’s an example:

def are_binaries_anagrams(num1, num2): bin1 = sorted(bin(num1)[2:]) bin2 = sorted(bin(num2)[2:]) return bin1 == bin2 # Example usage: print(are_binaries_anagrams(5, 6))

Output:

True

This code snippet defines a function `are_binaries_anagrams`

that takes two integers as arguments. It then converts each number to its binary representation with `bin()`

, strips the leading ‘0b’, and sorts the characters. This sorted version is then compared. As we see in the output, the binary strings for 5 and 6 (‘101’ and ‘110’) are indeed anagrams because it returns `True`

.

## Method 2: Counting Frequency

The frequency counting approach looks at the number of times each character (‘0’ or ‘1’) appears in the binary representation. We compare the counts for both numbers; if they match for both ‘0’ and ‘1’, the binary representations are anagrams. It’s efficient because we don’t need to sort the entire string, just count occurrences.

Here’s an example:

def binary_anagram_frequency(num1, num2): bin1, bin2 = bin(num1)[2:], bin(num2)[2:] return bin1.count('1') == bin2.count('1') and bin1.count('0') == bin2.count('0') # Example usage: print(binary_anagram_frequency(5, 6))

Output:

True

The function `binary_anagram_frequency`

converts the inputs to binary and counts the occurrences of ‘1’s and ‘0’s in each. With this approach, the need to sort is eliminated, improving efficiency.

## Method 3: Using Counter from Collections

The Counter class from Python’s collections module can be used to tally the occurrences of each character in the binary strings. The anagram check is a simple comparison of two Counter objects. This method is both concise and efficient, as the Counter object does the frequency calculation in a single step.

Here’s an example:

from collections import Counter def binary_anagram_counter(num1, num2): return Counter(bin(num1)[2:]) == Counter(bin(num2)[2:]) # Example usage: print(binary_anagram_counter(5, 6))

Output:

True

By using the `Counter`

class, `binary_anagram_counter`

performs an efficient comparison of the frequency of bits in the binary representations of the numbers, resulting in a compact and readable code.

## Method 4: Bit Manipulation

Bit manipulation methods can be used to determine if two numbers have the same number of ‘1’s in their binary representations without converting them to strings. This approach leverages the property that each bit shift and subtraction gets us closer to knowing the number of set bits.

Here’s an example:

def bit_count(n): count = 0 while n: n &= n - 1 count += 1 return count def binary_anagram_bitwise(num1, num2): return bit_count(num1) == bit_count(num2) # Example usage: print(binary_anagram_bitwise(5, 6))

Output:

True

This code leverages bitwise operations to count the number of ‘1’s in the binary representation of the numbers without converting them to strings. The `bit_count`

function reduces the number to 0 by continuously flipping the least significant ‘1’ bit and incrementing a counter.

## Bonus One-Liner Method 5: Functional Approach

This one-liner combines lambda functions with the principles from the previous methods to provide a functional and concise way to determine if binary representations of two numbers are anagrams or not.

Here’s an example:

# Functional one-liner binary_anagram_one_liner = lambda num1, num2: sorted(bin(num1)[2:]) == sorted(bin(num2)[2:]) # Example usage: print(binary_anagram_one_liner(5, 6))

Output:

True

The one-liner uses a lambda function to return the equality check of sorted binary strings of the two numbers β a clean and compact method that sticks to Python’s philosophy of being concise yet readable.

## Summary/Discussion

**Method 1:**Sort and Compare. Easy to understand. Not the most efficient due to sorting.**Method 2:**Counting Frequency. More efficient as it avoids sorting. May not be as readable as other methods.**Method 3:**Using Counter from Collections. Efficient and readable. Requires extra knowledge of the collections module.**Method 4:**Bit Manipulation. Most efficient as it operates directly with bit-level manipulation. Can be difficult to understand for beginners.**Method 5:**Functional Approach. Concise one-liner using the same logic as Method 1, but can be less intuitive for some readers.