💡 Problem Formulation: The challenge is to determine if two strings are “buddy strings.” Buddy strings are two strings of equal length where swapping just two characters in one of the strings will make it equal to the other string. For instance, given the input strings “ab” and “ba”, a swap of ‘a’ and ‘b’ in either string will result in the two matching, thus they are buddy strings.
Method 1: Brute Force Approach
This method involves checking each possible pair of indices and swapping characters at those positions to see if we can transform the first string into the second string. It’s a straightforward approach but not very efficient as it has O(n^2) complexity, where n is the string length.
Here’s an example:
def are_buddy_strings(s1, s2):
if len(s1) != len(s2):
return False
for i in range(len(s1)):
for j in range(i + 1, len(s1)):
if list(s1[:i] + s1[j] + s1[i + 1:j] + s1[i] + s1[j + 1:]) == list(s2):
return True
return False
print(are_buddy_strings("ab", "ba"))Output:
True
This function checks each index pair, swaps and compares the resultant string to the second string. It returns True when the pair of indices that makes the swap create strings that match.
Method 2: Optimized Swap Check
In this optimized version, we check for the conditions that would qualify strings as buddy strings more systematically, reducing the complexity. We look for pairs of differing characters and ensure there are at most two such, and they can be swapped.
Here’s an example:
def are_buddy_strings_optimized(s1, s2):
if len(s1) != len(s2) or sorted(s1) != sorted(s2):
return False
return sum(a != b for a, b in zip(s1, s2)) == 2
print(are_buddy_strings_optimized("ab", "ba"))Output:
True
The optimized function first ensures the strings are the same length and contain the same characters. Then, it only checks if there are exactly two pairs of different characters, meaning one swap can make the strings equal.
Method 3: Using a Counter
This method involves utilizing Python’s collections.Counter to count character frequencies. If the two strings have identical character counts, we then check if there are two characters to swap, or if the strings are the same and contain at least one duplicate character.
Here’s an example:
from collections import Counter
def are_buddy_strings_counter(s1, s2):
if len(s1) != len(s2) or Counter(s1) != Counter(s2):
return False
return sum(a != b for a, b in zip(s1, s2)) == 2 or (s1 == s2 and len(s1) - len(set(s1)) > 0)
print(are_buddy_strings_counter("ab", "ba"))Output:
True
This function utilizes Counter to quickly check for the same frequency of characters between the strings and then, similar to the optimized approach, verifies if there are two indices to swap or if the strings have duplicate characters and are equal.
Method 4: Character Position Mapping
This method maps the indices of different characters between the two strings. If there are only two such indices and a swap results in equal strings, we know they’re buddy strings. This is quite efficient for large strings with few differences.
Here’s an example:
def are_buddy_strings_mapping(s1, s2):
if len(s1) != len(s2):
return False
diff = [i for i in range(len(s1)) if s1[i] != s2[i]]
if len(diff) == 2 and s1[diff[0]] == s2[diff[1]] and s1[diff[1]] == s2[diff[0]]:
return True
return False
print(are_buddy_strings_mapping("ab", "ba"))Output:
True
This code snippet uses list comprehension to find indices where the strings differ. If there are exactly two such indices and swapping the characters at these indices between the strings results in matching strings, it returns True.
Bonus One-Liner Method 5: Compact Conditional
The one-liner encases the optimal method in a single, compact expression using a conditional. It’s a concise way to express the problem, though possibly at the cost of readability.
Here’s an example:
are_buddy_strings_one_liner = lambda s1, s2: len(s1) == len(s2) and ((len(set(zip(s1, s2))) == len(s1) - 1 and any(s1.count(c) > 1 for c in set(s1))) or len(set(zip(s1, s2))) == len(s1) + 1)
print(are_buddy_strings_one_liner("ab", "ba"))Output:
True
This one-liner uses set and zip to identify pairs of characters and then applies a conditional to determine if there are duplicate characters (allowing a swap) or if there are two characters that could be swapped based on the length comparison.
Summary/Discussion
- Method 1: Brute Force Approach. Tests all index pairs to find a match. Simple to understand but inefficient with large strings due to O(n^2) complexity.
- Method 2: Optimized Swap Check. Reduces steps by checking character sorting and the number of differing character pairs. Much faster than brute force, especially for large strings with few differences.
- Method 3: Using a Counter. Leverages the Counter class for quick frequency checking. It’s very readable and can be efficient with strings that have many repeated characters.
- Method 4: Character Position Mapping. Maps the position of differing characters, checking for possible swaps. Efficient but requires understanding list comprehension and conditional logic.
- One-Liner Method 5: Compact Conditional. A concise one-liner that contains the logic in a single expression. Although not the most readable, it is clever and reduces code lines.