Understanding the String Swap Problem

You are given two strings s1 and s2 of equal length. A string swap is an operation where you choose two indices in a string (not necessarily different) and swap the characters at these indices.

Return true if it is possible to make both strings equal by performing at most one string swap on exactly one of the strings. Otherwise, return false.

Examples of the String Swap Operation

Example 1:

Input: s1 = "bank", s2 = "kanb"
Output: true
Explanation: For example, swap the first character with the last character of s2 to make "bank".

Example 2:

Input: s1 = "attack", s2 = "defend"
Output: false
Explanation: It is impossible to make them equal with one string swap.

Example 3:

Input: s1 = "kelb", s2 = "kelb"
Output: true
Explanation: The two strings are already equal, so no string swap operation is required.

Constraints:

• 1 <= s1.length, s2.length <= 100
• s1.length == s2.length
• s1 and s2 consist of only lowercase English letters.

Optimal C++ Solution for Checking String Swaps

When implementing the algorithm to evaluate whether a single string swap can resolve the differences, the most efficient approach is to identify all differing indices. Using a modern programming language like C++, we can easily trace the divergent characters.

class Solution {
public:
    bool areAlmostEqual(string s1, string s2) {
        vector<int> index;
        for( uint8_t i = 0; i < s1.length(); i++ ){
            if( s1[i] != s2[i] )
                index.push_back(i);
            if( index.size() > 2)
                return false;
        }
        if( index.size() == 2 ){
            int i = index[0];
            int j = index[1];
            if( s1[i] == s2[j] && s1[j] == s2[i] ) return true;
        }
    
        return index.size() == 0;
    }
};

Semantic Breakdown and Algorithm Explanation

The code provided offers a clear method to check if one string swap can make strings equal. The core concept revolves around counting exactly how many characters do not match.

Step-by-Step Execution

• Initialization: A vector named index is initialized to store the positions of any mismatched characters between the two strings.
• Iteration: A loop runs through the entire length of the strings. Since the problem guarantees that both strings are of equal length, one loop serves perfectly to iterate through corresponding indices.
• Mismatch Detection: If a character at a specific position in s1 does not equal the character at the identical position in s2, that position is recorded.
• Early Exit Condition: If more than two mismatches are identified, the method instantly returns false. It is mathematically impossible to rectify more than two differing characters with a single swap.
• Validation: If exactly two mismatches exist, the algorithm inspects if a cross-swap resolves the issue. It checks if the first differing character in s1 matches the second differing character in s2, and vice-versa.

Complexity Analysis and Performance Evaluation

When solving algorithmic string challenges, evaluating performance is critical to establishing robust software systems. This solution is highly optimized for modern computing environments.

Time Complexity: O(N)

The time complexity is strictly linear, denoted as O(N), where N represents the total number of characters in the provided strings. The logic dictates exactly one continuous pass through the strings. The early exit condition (triggering a return when the mismatch count exceeds two) further guarantees that unnecessary iterations are completely avoided, establishing excellent average-case performance.

Space Complexity: O(1)

The space complexity is exceptionally lean at O(1) auxiliary space. We employ a dynamic array (the C++ vector) to store the indices of the mismatched elements. Because the early exit condition restricts this container from ever holding more than two integer elements, the memory footprint remains constant, irrespective of whether the input strings contain ten characters or ten million characters. This memory efficiency is a hallmark of high-quality software engineering.

Frequently Asked Questions (FAQ)

How do you check if one string swap makes two strings equal?

To check if one string swap makes two strings equal, compare both strings character by character. If there are exactly two differing positions, ensure that swapping these characters in one string matches the other string. If there are zero differences, no swaps are needed. Any other number of differences means they cannot be made equal with one swap.

What is the time complexity of the string swap check algorithm?

The time complexity is O(N), where N is the length of the strings, because the algorithm only needs to iterate through the strings once to find differences. The space complexity is O(1) auxiliary space, as we at most store two indices.