**π‘ Problem Formulation:** This article addresses the challenge of determining the shortest binary string possible after continually removing pairs of adjacent bits that are different. For example, given an input like `"10101"`

, the desired output after performing the operation would be `"1"`

, since we can remove “10” or “01” pairs, ultimately ending with a single character.

## Method 1: Iterative Pair Removal

This first method involves iterating through the string and repeatedly removing adjacent pairs that are different. The operation continues until no more removable pairs exist. The function specifically looks for ’10’ or ’01’ pairs and deletes them until the string cannot be reduced further.

Here’s an example:

def iterative_pair_removal(s): stack = [] for bit in s: if stack and stack[-1] != bit: stack.pop() else: stack.append(bit) return ''.join(stack) print(iterative_pair_removal("10101"))

Output: `"1"`

This code snippet uses a stack to effectively keep track of bits which can be paired and removed. Each bit is compared with the top of the stack. If they’re different, the top bit is popped (pair removed); otherwise, the bit is stacked. This mimics removal of pairs and halts when no more pairs can be formed.

## Method 2: Using Regular Expressions

The second method takes advantage of Python’s powerful `re`

module to replace ’10’ or ’01’ pairs with an empty string in a loop until no further changes can be made.

Here’s an example:

import re def regex_pair_removal(s): while re.search(r'(10|01)', s): s = re.sub(r'(10|01)', '', s) return s print(regex_pair_removal("10101"))

Output: `"1"`

The snippet creates a loop that continues to search and replace occurrences of ’10’ or ’01’ with an empty string until it can’t find any more matches, leading to the shortest possible string.

## Method 3: Two-Pointer Technique

This method leverages a two-pointer technique, iterating over the string with a fast and slow pointer to identify removable pairs and skip over them, slowly building the resultant string.

Here’s an example:

def two_pointer_removal(s): result = [] i = 0 while i < len(s): if result and result[-1] != s[i]: result.pop() else: result.append(s[i]) i += 1 return ''.join(result) print(two_pointer_removal("10101"))

Output: `"1"`

The code uses two pointers, although not explicitly, with one implied by the length of the `result`

list. When a non-matching bit is found, it’s appended, else the last element is popped out, effectively removing the adjacent pairs.

## Method 4: Recursive Reduction

Method 4 relies on recursion to remove adjacent pairs of differing bits. If a removable pair is found, the string is reduced and the recursive function is called until no more removable pairs exist.

Here’s an example:

def recursive_pair_removal(s): for i in range(len(s) - 1): if s[i] != s[i + 1]: return recursive_pair_removal(s[:i] + s[i+2:]) return s print(recursive_pair_removal("10101"))

Output: `"1"`

This recursion pattern shortens the string by cutting out pairs at each recursive call. The function proceeds until it traverses a pass without finding any pairs to remove, indicating it has reached the shortest possible string.

## Bonus One-Liner Method 5: Reduce Function with Lambda

Employing Python’s `functools.reduce()`

function and a concise lambda, this one-liner solution accumulates each bit onto a stack, popping only when a removable pair is encountered.

Here’s an example:

from functools import reduce shortest_string = reduce(lambda acc, x: acc[:-1] if acc[-1:] != acc and acc[-1:] != x else acc + x, "10101", "") print(shortest_string)

Output: `"1"`

This elegant one-liner uses the `reduce()`

function to effectively construct a stack that gains or loses elements based on the preceding character, echoing the removal of ’10’ or ’01’ pairs.

## Summary/Discussion

**Method 1:**Iterative Pair Removal. Easy to understand. Can be slower for long strings due to repeated list operations.**Method 2:**Using Regular Expressions. Powerful for pattern matching. May not be as intuitive for those unfamiliar with regex. Can become expensive with very large strings.**Method 3:**Two-Pointer Technique. Space efficient and intuitive. Iterating over every element can lead to higher time complexity with very long strings.**Method 4:**Recursive Reduction. Conceptually simple recursive pattern. The risk of hitting recursion depth limits with very long strings, and higher overhead due to function calls.**Method 5:**Reduce Function with Lambda. Extremely concise. Might be less readable to those not familiar with functional programming concepts.