**π‘ Problem Formulation:** We aim to find the minimum number of operations required to remove adjacent pairs of identical bits from a binary string. Consider a binary string “110011”. The desired output after performing operations to remove consecutive identical bits would be “0” with the number of operations being three (removing ’11’ twice and ’00’ once).

## Method 1: Iterative Comparison

This method employs a simple iterative comparison between adjacent elements in a binary string. Each time a pair of consecutive identical bits is found, they are removed, and the operations counter is incremented.

Here’s an example:

def remove_consecutive_bits_iteratively(binary_string): operations = 0 i = 0 while i < len(binary_string) - 1: if binary_string[i] == binary_string[i + 1]: binary_string = binary_string[:i] + binary_string[i+2:] operations += 1 i = max(0, i - 1) else: i += 1 return operations print(remove_consecutive_bits_iteratively("110011"))

Output: `3`

This code snippet defines the function `remove_consecutive_bits_iteratively()`

, which takes a binary string as its input and returns the count of operations needed to remove consecutive identical bits. It uses a while loop to traverse the string, and whenever it finds two consecutive identical bits, it slices the string to remove them while also decrementing the index to check for new consecutive bits that might have formed before proceeding.

## Method 2: Using Stack

In this method, a stack data structure is used to keep track of the bits. As we iterate through the binary string, if the current bit is different from the top of the stack, it is pushed; otherwise, the top bit is popped (removed), indicating the removal of a pair of consecutive identical bits, and an operation is counted.

Here’s an example:

def remove_consecutive_bits_stack(binary_string): stack = [] operations = 0 for bit in binary_string: if stack and stack[-1] == bit: stack.pop() operations += 1 else: stack.append(bit) return operations print(remove_consecutive_bits_stack("110011"))

Output: `3`

The `remove_consecutive_bits_stack()`

function uses a stack to process the binary string. If the current bit is the same as the one on the stack’s top, it indicates a pair of identical consecutive bits, so the top bit is popped off, and we increment the operations counter. If they’re different, the current bit is pushed onto the stack instead.

## Method 3: Regular Expression Substitution

Regular expressions provide an elegant and compact way to match patterns. This method uses a regular expression to find and perform substitutions on all occurrences of consecutive identical bits until no more such pairs exist, counting each substitution as an operation.

Here’s an example:

import re def remove_consecutive_bits_regex(binary_string): operations = 0 pattern = r'(00|11)' while re.search(pattern, binary_string): binary_string = re.sub(pattern, '', binary_string, 1) operations += 1 return operations print(remove_consecutive_bits_regex("110011"))

Output: `3`

The `remove_consecutive_bits_regex()`

function uses Python’s `re`

module to repeatedly apply a regular expression substitution that targets pairs of ’00’ or ’11’. Whenever a substitution is made, which signifies a removal operation, the operation counter is increased by one.

## Method 4: Recursive Approach

The recursive approach breaks down the problem into smaller instances of the same problem. It recursively removes consecutive identical bits by deleting the first pair found and then calling itself on the newly formed string until no more such pairs exist.

Here’s an example:

def remove_consecutive_bits_recursive(binary_string): for i in range(len(binary_string) - 1): if binary_string[i] == binary_string[i + 1]: return 1 + remove_consecutive_bits_recursive(binary_string[:i] + binary_string[i+2:]) return 0 print(remove_consecutive_bits_recursive("110011"))

Output: `3`

This recursive function, `remove_consecutive_bits_recursive()`

, iterates through the binary string looking for a pair of consecutive identical bits. When such a pair is found, it is removed, and the function is called again on the remaining string. The base case is reached when no such pair is found, and the recursion ends.

## Bonus One-Liner Method 5: List Comprehension with zip()

This one-liner uses Python’s list comprehension in tandem with the `zip()`

function to count the number of consecutive pairs by comparing each bit with its successor.

Here’s an example:

def remove_consecutive_bits_one_liner(binary_string): return sum(1 for a, b in zip(binary_string, binary_string[1:]) if a == b) print(remove_consecutive_bits_one_liner("110011"))

Output: `3`

The function `remove_consecutive_bits_one_liner()`

counts the number of consecutive pairs using `zip()`

to create pairs of adjacent bits from the string. The list comprehension iterates over these pairs and increments the count whenever a pair of identical bits is found.

## Summary/Discussion

**Method 1:**Iterative Comparison. Simple and straight-forward algorithm. However, it might be inefficient for very long strings due to repeated string slicing operations.**Method 2:**Using Stack. More space-efficient than method 1 and avoids frequent string manipulations. Good performance on longer strings. However, stack size could grow depending on input.**Method 3:**Regular Expression Substitution. Compact and utilizes powerful regex operations. May not be as efficient as other methods for extremely long strings due to regex complexity.**Method 4:**Recursive Approach. Clean logic and easy to understand. However, for very large inputs, it might cause a stack overflow due to deep recursion levels.**Bonus One-Liner Method 5:**List Comprehension with`zip()`

. Short and elegant, but under the hood, it still creates temporary lists which could be a memory overhead for large inputs.