algoadvance

Given two strings s1 and s2 of equal length, your task is to determine if there is a way to swap exactly one pair of characters in s1 so that it becomes equal to s2.

Clarifying Questions

1. Input Constraints:
   • Are s1 and s2 guaranteed to have the same length?
   • What is the maximum length of s1 and s2?
2. Output:
   • Should the function return a boolean value indicating whether a single swap can make the strings equal?

Strategy

To solve this problem, follow these steps:

1. Comparison: First, ensure s1 and s2 have equal lengths. If they don’t, return False immediately (given the constraints, though, this should not happen).

2. Identifying Mismatch: Traverse both strings and identify the indices where they differ.

3. Conditions for Swap:

   • If the strings are already equal, a swap is not necessary, but technically a “no-op swap” can be considered valid.
   • If there are exactly two positions where the characters in s1 and s2 differ, check if swapping these two positions in s1 will equalize the two strings.
   • If there are more or less than two mismatches, it is impossible to make the strings equal with just one swap.

Code

def areAlmostEqual(s1: str, s2: str) -> bool:
    # Step 1: Check if lengths of s1 and s2 are the same
    if len(s1) != len(s2):
        return False

    # Step 2 & 3: Find all positions where the strings differ
    mismatches = []
    for i in range(len(s1)):
        if s1[i] != s2[i]:
            mismatches.append(i)

    # Step 4: Evaluate mismatch conditions
    if len(mismatches) == 0:
        # Strings are already equal; one "no-op swap" is trivially possible
        return True
    elif len(mismatches) == 2:
        # Strings differ at exactly two positions
        i, j = mismatches
        # Check if swapping s1[i] with s1[j] will match s2
        if s1[i] == s2[j] and s1[j] == s2[i]:
            return True
    
    # In all other cases, return False
    return False

Time Complexity

• Time Complexity: The algorithm involves a single traversal of both strings, thus it operates in O(n) time, where n is the length of the strings.
• Space Complexity: The additional space used is simply for storing the mismatched indices, which in the worst case could be O(n), but typically we only need to store up to 2 indices. Thus, average case space complexity is O(1), and the worst case is O(n).

By following this strategy, you can effectively determine if the strings s1 and s2 can be made equal with exactly one swap.

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