Problem Description

You are given two strings s1 and s2 that have the same length. A string swap operation allows you to select any two indices in a string (they can be the same index) and exchange the characters at those positions.

Your task is to determine if you can make the two strings identical by performing at most one swap operation on exactly one of the two strings.

The function should return true if it's possible to make both strings equal with this constraint, and false otherwise.

For example:

• If s1 = "bank" and s2 = "kanb", you can swap indices 0 and 3 in s2 to get "bank", making both strings equal. This would return true.
• If s1 = "attack" and s2 = "defend", no single swap can make them equal, so this would return false.
• If s1 = "kelb" and s2 = "kelb", the strings are already equal (zero swaps needed), so this would return true.

The key insight is that for two strings to be made equal with at most one swap:

1. They must either be already identical (0 differences)
2. Or have exactly 2 positions where characters differ, and swapping those two characters in one string would make them match the other string

Intuition

When we think about making two strings equal with at most one swap operation on exactly one string, we need to consider what patterns would allow this to happen.

First, let's think about the possible scenarios:

• If the strings are already identical, we need 0 swaps, which satisfies "at most one swap"
• If the strings differ at exactly 2 positions, we might be able to fix them with 1 swap
• If the strings differ at 1 position or more than 2 positions, it's impossible to make them equal with just one swap

The key realization is that when we perform a swap operation, we're changing exactly 2 characters in one string (or 0 if we swap a position with itself). This means a single swap can only fix differences at exactly 2 positions.

For the strings to become equal after one swap, the mismatched characters must be "complementary." If at position i we have s1[i] = 'a' and s2[i] = 'b', and at position j we have s1[j] = 'b' and s2[j] = 'a', then swapping positions i and j in either string would make them equal.

This leads us to track:

1. How many positions have different characters (cnt)
2. What are the mismatched characters at the first difference (c1 and c2)
3. When we encounter the second difference, check if the characters are the "reverse" of the first difference

The elegant part is that we don't need to store all differences - we only care about whether there are 0, 2, or some other number of differences. If there are exactly 2 differences and they're complementary (can be fixed by one swap), we return true. Otherwise, we return false.

Solution Approach

The solution uses a counting approach with early termination to efficiently determine if two strings can be made equal with at most one swap.

We initialize:

• cnt = 0: Counter for the number of positions where characters differ
• c1 = None and c2 = None: Variables to store the first pair of mismatched characters

We traverse both strings simultaneously using zip(s1, s2). For each pair of characters (a, b) at the same position:

1. When characters differ (a != b):

   • Increment cnt by 1
   • Check for immediate failure conditions:
     • If cnt > 2: More than 2 differences means we can't fix with one swap, return False
     • If cnt == 2 and the second mismatch doesn't complement the first:
       • Check if a != c2 or b != c1
       • This verifies whether swapping would actually fix both mismatches
       • If not complementary, return False
   • Store the mismatched characters: c1 = a and c2 = b

2. After traversal completes:

   • Return cnt != 1
   • This handles three valid cases:
     • cnt == 0: Strings are already equal (no swap needed)
     • cnt == 2: Exactly two differences that are complementary (validated during traversal)
   • And rejects the invalid case:
     • cnt == 1: Only one difference can't be fixed with a swap

The algorithm runs in O(n) time where n is the length of the strings, and uses O(1) extra space. The early termination when cnt > 2 ensures we don't waste time checking the rest of the strings once we know they can't be made equal with one swap.

Example Walkthrough

Let's walk through the solution with s1 = "bank" and s2 = "kanb".

Initial State:

• cnt = 0 (no differences found yet)
• c1 = None, c2 = None (no mismatched characters stored)

Step-by-step traversal:

1. Position 0: s1[0] = 'b', s2[0] = 'k'

   • Characters differ, so cnt = 1
   • Store the mismatch: c1 = 'b', c2 = 'k'
   • Continue (only 1 difference so far)

2. Position 1: s1[1] = 'a', s2[1] = 'a'

   • Characters match, no action needed
   • cnt remains 1

3. Position 2: s1[2] = 'n', s2[2] = 'n'

   • Characters match, no action needed
   • cnt remains 1

4. Position 3: s1[3] = 'k', s2[3] = 'b'

   • Characters differ, so cnt = 2
   • Check if this second mismatch complements the first:
     • Current: s1[3] = 'k', s2[3] = 'b'
     • First mismatch: c1 = 'b', c2 = 'k'
     • Is s1[3] == c2? Yes, 'k' == 'k' ✓
     • Is s2[3] == c1? Yes, 'b' == 'b' ✓
   • The mismatches are complementary! If we swap positions 0 and 3 in s2, we get:
     • s2 becomes "bank" (swapping 'k' and 'b')
   • Continue traversal

Final Check:

• cnt = 2
• Return cnt != 1 → 2 != 1 → True

The algorithm correctly identifies that these strings can be made equal with exactly one swap operation (swapping indices 0 and 3 in s2).

Counter-example: If we had s1 = "abcd" and s2 = "abdc", we'd find differences at positions 2 and 3. However, s1[2] = 'c', s2[2] = 'd' and s1[3] = 'd', s2[3] = 'c'. These are complementary (swapping positions 2 and 3 in either string makes them equal), so the function returns True.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the strings s1 and s2. The algorithm iterates through both strings exactly once using the zip function, performing constant-time operations (comparisons and variable assignments) for each character pair.

The space complexity is O(1). The algorithm uses only a fixed amount of extra space regardless of the input size: one counter variable cnt and two variables c1 and c2 to store characters. These variables do not scale with the input size.

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

Common Pitfalls

1. Forgetting to Handle the Already-Equal Case

A common mistake is focusing only on finding exactly 2 differences and forgetting that strings that are already identical should also return true.

Incorrect approach:

def areAlmostEqual(self, s1: str, s2: str) -> bool:
    # ... finding differences ...
    return diff_count == 2  # Wrong! Misses the case where diff_count == 0

Solution: Always check for diff_count == 0 OR diff_count == 2, which is elegantly handled by return diff_count != 1.

2. Not Validating the Swap Would Actually Work

Another pitfall is counting exactly 2 differences but not verifying that swapping those two positions would actually make the strings equal.

Incorrect approach:

def areAlmostEqual(self, s1: str, s2: str) -> bool:
    diff_count = 0
    for char_s1, char_s2 in zip(s1, s2):
        if char_s1 != char_s2:
            diff_count += 1
    return diff_count == 0 or diff_count == 2  # Wrong! Doesn't verify the swap works

For example, with s1 = "ab" and s2 = "cd", this would incorrectly return true even though swapping can't make them equal.

Solution: Store the first pair of differing characters and verify that the second pair is exactly the reverse (i.e., char_s1 == first_char_s2 and char_s2 == first_char_s1).

3. Inefficient String Comparison After Performing Swaps

Some solutions actually perform the swap operation and then compare entire strings, which is less efficient.

Inefficient approach:

def areAlmostEqual(self, s1: str, s2: str) -> bool:
    if s1 == s2:
        return True
  
    for i in range(len(s1)):
        for j in range(i, len(s1)):
            # Actually perform swap on s1
            s1_list = list(s1)
            s1_list[i], s1_list[j] = s1_list[j], s1_list[i]
            if ''.join(s1_list) == s2:
                return True
    return False

This has O(n³) time complexity due to nested loops and string comparisons.

Solution: Use the counting approach with early termination as shown in the original solution, achieving O(n) time complexity.