**π‘ Problem Formulation:** You are given two strings of equal length and the objective is to determine if one string can be made equal to the other by swapping characters within the strings. For example, the string `"converse"`

can be turned into `"conserve"`

by swapping the ‘v’ and ‘s’ characters.

## Method 1: Brute Force Swap

This method involves trying every possible swap between the two strings until they are equal or it’s determined that it isn’t possible. A recursive function is typically used to try all permutations of swaps. While simple, this approach is not efficient and is not recommended for long strings due to its exponential time complexity.

Here’s an example:

def can_be_equal_by_swapping(str1, str2): if str1 == str2: return True for i in range(len(str1)): for j in range(i + 1, len(str2)): if can_be_equal_by_swapping(str1[:i] + str2[j] + str1[i+1:j] + str2[i] + str1[j+1:], str2): return True return False # Example usage: result = can_be_equal_by_swapping("converse", "conserve") print(result)

The output would be:

`True`

This snippet defines a recursive function that attempts to swap each pair of characters in the string until either the strings are equal or all possibilities have been exhausted. While this code covers all scenarios, it may lead to a stack overflow for long strings due to the depth of recursion.

## Method 2: Greedy Swap with Hashing

A more efficient approach than brute force is to use a greedy strategy complemented by hashing. This method works only if the characters that need to be swapped are unique. It involves finding the positions of the mismatched characters and swapping them if possible. Hash maps (dictionaries) are used for fast lookups of character indices.

Here’s an example:

def can_swap_to_equal(str1, str2): if str1 == str2: return True else: index_map = {} for index, char in enumerate(str1): if char in index_map: return False index_map[char] = index for index, char in enumerate(str2): if char not in index_map: return False str1 = list(str1) str1[index_map[char]], str1[index] = str1[index], str1[index_map[char]] if ''.join(str1) == str2: return True return False # Example usage: result = can_swap_to_equal("converse", "conserve") print(result)

The output would be:

`True`

This snippet uses a greedy algorithm and a hash map to find and perform necessary swaps. This approach minimizes the number of swaps needed to determine equality. Still, it assumes unique characters for swaps, which may not always be applicable.

## Method 3: Character Count Comparison

In this method, we perform a count of each character in both strings and compare the counts. If they are the same, the strings can be made equal by swapping. This is the most efficient method for determining if a swap is possible, working well for long strings with a limited set of characters.

Here’s an example:

from collections import Counter def can_transform(str1, str2): return Counter(str1) == Counter(str2) # Example usage: result = can_transform("converse", "conserve") print(result)

The output would be:

`True`

The code utilizes Python’s `collections.Counter`

class to quickly count the characters in both strings and compares the resulting dictionaries. This method is fast and works well with repeated characters but does not give any information about the swap operations needed.

## Method 4: Optimized Swap Checker

This optimized method looks at each mismatched pair of characters. If there is a way to swap the characters within the other string to make the pair match, then we proceed, otherwise, it’s impossible. This method leverages two-pointers to reduce the complexity.

Here’s an example:

def possible_to_swap(str1, str2): if len(str1) != len(str2): return False pairs = [] for a, b in zip(str1, str2): if a != b: pairs.append((a, b)) if not pairs: return True # strings are already equal return len(pairs) == 2 and pairs[0] == pairs[1][::-1] # Example usage: result = possible_to_swap("converse", "conserve") print(result)

The output would be:

`True`

This code snippet efficiently determines if two strings can be made equal with a single pair of swaps. It creates a list of mismatched character pairs and checks if swapping a pair in one of the strings can resolve the differences. This both minimizes operations and provides a quick answer.

## Bonus One-Liner Method 5: Using XOR

For a fun, though not always practical application, one can use the bitwise XOR operation to check if the strings can be equalized by character swaps, under very specific conditions where character uniqueness and string encoding allow such operations.

Here’s an example:

def can_equal_by_xor(str1, str2): return not any(ord(a) ^ ord(b) for a, b in zip(str1, str2)) # Example usage: result = can_equal_by_xor("ac", "ca") print(result)

The output would be:

`True`

This code snippet applies the bitwise XOR operation for each pair of corresponding characters in the two strings and checks if any result is non-zero. However, this method is just a clever trick devoid of practical applications for general-case string swapping problems.

## Summary/Discussion

**Method 1: Brute Force Swap.**Simple to understand. Not efficient for longer strings due to exponential time complexity.**Method 2: Greedy Swap with Hashing.**Efficient for unique character swaps. Not suitable for duplicate characters in strings.**Method 3: Character Count Comparison.**Efficient for even long strings with limited character set. Does not provide information on swaps needed.**Method 4: Optimized Swap Checker.**Provides a quick answer. Efficient with two-pointers strategy. Works only if a single swap can make the strings equal.**Bonus Method 5: Using XOR.**A one-liner that’s more of a programming novelty. Not generally applicable for solving the string swap problem.