Problem Description

You are given two strings s and goal. You need to determine if you can make s equal to goal by swapping exactly two letters in s.

The swap operation involves selecting two different positions i and j in string s (where i != j) and exchanging the characters at those positions.

The function should return true if exactly one swap can transform s into goal, and false otherwise.

Key Points:

• You must perform exactly one swap (exchanging two characters)
• The two positions being swapped must be different (i != j)
• If s is already equal to goal, you still need to perform a swap - this is only valid if there are duplicate characters that can be swapped without changing the string

Examples:

• If s = "ab" and goal = "ba", swapping positions 0 and 1 gives "ba", so return true
• If s = "ab" and goal = "ab", they are already equal but there are no duplicate characters to swap, so return false
• If s = "aa" and goal = "aa", they are equal and we can swap the two 'a's without changing the string, so return true
• If s = "abcd" and goal = "cbad", swapping positions 0 and 2 transforms "abcd" to "cbad", so return true

Intuition

To solve this problem, let's think about what conditions must be met for exactly one swap to transform s into goal.

First, both strings must have the same length - if they don't, no single swap can make them equal.

Second, both strings must contain the same characters with the same frequencies. A swap only rearranges characters within s, it doesn't add or remove any. So if s has different character counts than goal, they can never be made equal through swapping.

Now, assuming the strings have the same length and character frequencies, we need to consider two scenarios:

Scenario 1: The strings differ at exactly 2 positions If s and goal differ at exactly 2 positions, say positions i and j, then swapping s[i] with s[j] should make them equal. For this to work, we need s[i] = goal[j] and s[j] = goal[i]. The character frequency check we already performed guarantees this condition.

Scenario 2: The strings are already identical If s and goal are already the same, we still need to perform one swap. The only way this works is if there's at least one character that appears more than once in s. We can swap two occurrences of the same character, which doesn't change the string but satisfies the requirement of performing exactly one swap.

If the strings differ at 0 positions (identical) but have no duplicate characters, or if they differ at 1, 3, or more positions, then it's impossible to make them equal with exactly one swap.

This leads us to count the number of differing positions and check for duplicate characters when needed.

Solution Approach

Let's walk through the implementation step by step:

Step 1: Check Length Equality

m, n = len(s), len(goal)
if m != n:
    return False

First, we check if both strings have the same length. If they don't, it's impossible to make them equal with a single swap.

Step 2: Check Character Frequency

cnt1, cnt2 = Counter(s), Counter(goal)
if cnt1 != cnt2:
    return False

We use Python's Counter from the collections module to count the frequency of each character in both strings. If the character frequencies don't match, no amount of swapping within s can make it equal to goal. The Counter creates a dictionary-like object where keys are characters and values are their counts.

Step 3: Count Differences

diff = sum(s[i] != goal[i] for i in range(n))

We count how many positions have different characters between s and goal. This uses a generator expression with sum() to count the number of True values (positions where characters differ).

Step 4: Determine if Valid Swap Exists

return diff == 2 or (diff == 0 and any(v > 1 for v in cnt1.values()))

The final check handles two valid cases:

• diff == 2: Exactly two positions differ. Since we've already verified the character frequencies match, swapping these two positions will make the strings equal.
• diff == 0 and any(v > 1 for v in cnt1.values()): The strings are already identical, but we can still perform a valid swap if there's at least one character that appears more than once. We check this by looking at the values in our frequency counter to see if any character has a count greater than 1.

The algorithm has a time complexity of O(n) where n is the length of the strings, as we traverse the strings a constant number of times. The space complexity is O(k) where k is the number of unique characters in the strings (for the Counter objects).

Example Walkthrough

Let's walk through the solution with s = "abcd" and goal = "cbad":

Step 1: Check Length Equality

• len(s) = 4, len(goal) = 4 ✓
• Lengths match, continue to next step

Step 2: Check Character Frequency

• Counter(s) = {'a': 1, 'b': 1, 'c': 1, 'd': 1}
• Counter(goal) = {'c': 1, 'b': 1, 'a': 1, 'd': 1}
• Both counters are equal (same characters with same frequencies) ✓

Step 3: Count Differences

• Compare position by position:
  • Position 0: s[0] = 'a', goal[0] = 'c' → Different
  • Position 1: s[1] = 'b', goal[1] = 'b' → Same
  • Position 2: s[2] = 'c', goal[2] = 'a' → Different
  • Position 3: s[3] = 'd', goal[3] = 'd' → Same
• Total differences: diff = 2

Step 4: Determine if Valid Swap Exists

• Since diff == 2, we check if swapping positions 0 and 2 works:
  • s[0] = 'a' should become goal[0] = 'c'
  • s[2] = 'c' should become goal[2] = 'a'
  • After swapping positions 0 and 2: "abcd" → "cbad" ✓
• Return true

Let's also consider the edge case where s = "aa" and goal = "aa":

Steps 1-2: Length and frequency checks pass

Step 3: Count differences = 0 (strings are identical)

Step 4: Check if we can still perform a valid swap:

• diff == 0, so check if any character appears more than once
• Counter(s) = {'a': 2}, and 2 > 1 ✓
• We can swap the two 'a's at positions 0 and 1, which keeps the string unchanged
• Return true

Solution Implementation

Time and Space Complexity

Time Complexity: O(n)

The algorithm performs the following operations:

• Creating Counter(s): O(n) where n is the length of string s
• Creating Counter(goal): O(n) where n is the length of string goal
• Comparing two Counter objects cnt1 != cnt2: O(n) in worst case (needs to check all unique characters)
• Computing differences using list comprehension: O(n) to iterate through all indices
• Checking if any value > 1 in cnt1.values(): O(k) where k is the number of unique characters, and k ≤ n

Since all operations are linear and performed sequentially, the overall time complexity is O(n).

Space Complexity: O(k)

The algorithm uses additional space for:

• Counter(s): O(k₁) where k₁ is the number of unique characters in s
• Counter(goal): O(k₂) where k₂ is the number of unique characters in goal
• The generator expression for computing diff doesn't create additional space beyond O(1)

In the worst case where all characters are unique, k can be at most min(n, 26) for lowercase English letters. Since the alphabet size is constant (26 for lowercase letters), the space complexity can also be considered O(1) for practical purposes. However, if we consider the general case without alphabet constraints, the space complexity is O(k) where k ≤ n.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall 1: Forgetting the Case Where Strings Are Already Equal

The Problem: Many developers initially write code that only checks if there are exactly 2 differences between the strings, forgetting that when s and goal are already identical, we still need to perform exactly one swap. This leads to incorrect handling of cases like:

• s = "ab", goal = "ab" → Should return False (no duplicates to swap)
• s = "aa", goal = "aa" → Should return True (can swap the two 'a's)

Incorrect Implementation:

def buddyStrings(self, s: str, goal: str) -> bool:
    if len(s) != len(goal):
        return False
  
    differences = []
    for i in range(len(s)):
        if s[i] != goal[i]:
            differences.append(i)
  
    # WRONG: Only checking for 2 differences
    return len(differences) == 2

The Solution: Always handle both cases explicitly:

1. When there are exactly 2 differences
2. When there are 0 differences AND at least one duplicate character exists

# Correct approach
difference_count = sum(s[i] != goal[i] for i in range(len(s)))
return (difference_count == 2 or 
        (difference_count == 0 and any(count > 1 for count in Counter(s).values())))

Pitfall 2: Not Validating That the Two Differences Are Actually Swappable

The Problem: When finding exactly 2 differences, some implementations forget to verify that swapping those two positions would actually make the strings equal. Just having 2 differences doesn't guarantee a valid swap.

Incorrect Implementation:

def buddyStrings(self, s: str, goal: str) -> bool:
    if len(s) != len(goal):
        return False
  
    diff_positions = []
    for i in range(len(s)):
        if s[i] != goal[i]:
            diff_positions.append(i)
  
    # WRONG: Not checking if the swap actually works
    if len(diff_positions) == 2:
        return True  # This assumes any 2 differences can be swapped
  
    return False

The Solution: Either explicitly check that swapping the two different positions produces the correct result, OR use character frequency counting to ensure the strings contain the same characters:

# Solution 1: Explicit swap verification
if len(diff_positions) == 2:
    i, j = diff_positions[0], diff_positions[1]
    return s[i] == goal[j] and s[j] == goal[i]

# Solution 2: Character frequency validation (as in our main solution)
s_char_count = Counter(s)
goal_char_count = Counter(goal)
if s_char_count != goal_char_count:
    return False  # Different character sets means no valid swap exists

Pitfall 3: Using Set to Check for Duplicates When Strings Are Equal

The Problem: When checking if identical strings have duplicate characters (to allow a valid swap), using len(set(s)) < len(s) seems elegant but can be inefficient for very long strings with many unique characters.

Less Efficient Implementation:

if difference_count == 0:
    # Creates a full set of all characters
    return len(set(s)) < len(s)

The Solution: Use early termination with any() to stop as soon as a duplicate is found:

# More efficient - stops as soon as a duplicate is found
if difference_count == 0:
    return any(count > 1 for count in Counter(s).values())

This approach is particularly beneficial when dealing with strings that have duplicates early in the character frequency distribution, as it doesn't need to process all unique characters.