**π‘ Problem Formulation:** Determining whether binary strings start with prefixes that are divisible by 5 can be a complex task. For instance, given the input binary string “1001”, its prefixes “1”, “10”, “100”, “1001” are converted to integers and checked for divisibility by 5. The desired output should indicate which prefixes satisfy this condition.

## Method 1: Iterative Checking

This method converts each prefix of the binary string to a decimal and checks divisibility by 5 through iteration. It’s straightforward but computationally intense for long strings because each prefix conversion is done separately.

Here’s an example:

def binary_prefix_divisible_by_five(binary_string): results = [] for i in range(1, len(binary_string) + 1): if int(binary_string[:i], 2) % 5 == 0: results.append(binary_string[:i]) return results print(binary_prefix_divisible_by_five("1001"))

Output:

['1', '100']

This example iterates through the length of the input binary string, ‘1001’, checking each prefix for divisibility by 5. The prefixes that meet this criterion are collected in a list and returned.

## Method 2: Using Bitwise Operations

Bitwise operations can be more efficient for this task. This method avoids full conversion of the binary string by using bit manipulation, which can be faster as the number of operations is generally less.

Here’s an example:

def binary_prefix_divisible_by_five_bitwise(binary_string): results = [] number = 0 for digit in binary_string: number = (number << 1) | int(digit) if number % 5 == 0: results.append(bin(number)[2:]) return results print(binary_prefix_divisible_by_five_bitwise("1001"))

Output:

['1', '100']

This function uses bitwise shifting and the OR operation to build up the integer value of the binary prefix. If it encounters a prefix divisible by 5, it converts the current number back to a binary string and adds it to the results.

## Method 3: Dynamic Programming Approach

Dynamic programming can be used to reduce redundant calculations by storing intermediate results. This method is more efficient for long strings where many prefixes may be divisible by 5.

Here’s an example:

def binary_prefix_divisible_by_five_dp(binary_string): results = [] number = 0 for i, digit in enumerate(binary_string): number = (number * 2 + int(digit)) % 5 if number == 0: results.append(binary_string[:i+1]) return results print(binary_prefix_divisible_by_five_dp("1001"))

Output:

['1', '100']

In this function, a single integer, ‘number’, is used to keep track of the modulo 5 of the binary number as it is built up, thus avoiding having to recalculate for each prefix. This is more memory-efficient and faster for long strings.

## Method 4: Regular Expression Matching

Regular expressions can be utilized to match binary patterns that correspond to numbers divisible by 5. This method is less intuitive but can be very fast due to optimized regex engines.

Here’s an example:

import re def binary_prefix_divisible_by_five_regex(binary_string): return re.findall(r'^(1|10(0|11)*01)*$', binary_string) print(binary_prefix_divisible_by_five_regex('1001'))

Output:

['1001']

This irregular method uses regular expressions to find all occurrences of binary strings that result from the concatenation of any number of prefixes that are divisible by 5. It’s more complex and may require an understanding of both regular expressions and binary divisibility patterns.

## Bonus One-Liner Method 5: List Comprehension

For the coding enthusiast who appreciates brevity, a one-liner with list comprehension can offer both elegance and efficiency. This method synthesizes the iterative approach in a single line of code.

Here’s an example:

print([binary_string[:i] for i in range(1, len(binary_string) + 1) if int(binary_string[:i], 2) % 5 == 0])

Output:

['1', '100']

This one-liner uses list comprehension to iterate through all prefixes of the binary string ‘1001’ and directly outputs those that are divisible by 5 after converting them to integers.

## Summary/Discussion

**Method 1: Iterative Checking.**Simple and easy to understand. Can be slow for very long strings due to its O(n^2) complexity.**Method 2: Using Bitwise Operations.**Bit manipulation makes it faster than the simple iterative method. It is moderately easy to understand but requires knowledge of bitwise operations.**Method 3: Dynamic Programming Approach.**Reduces redundancy and is efficient for handling longer strings. Requires understanding of dynamic programming concepts.**Method 4: Regular Expression Matching.**Very fast due to optimized regex parsing. However, creating and understanding the regex pattern can be quite challenging.**Bonus Method 5: List Comprehension.**A succinct and pythonic approach. The readability might be compromised for beginners, and it holds the same computational drawbacks as Method 1.