The ‘Check if One String Swap Can Make Strings Equal’ is a captivating Leetcode problem that falls under the category of string manipulation and comparison. This problem is an excellent test of your understanding of basic string operations, conditional statements, and list comprehensions.

In this extensive article, we will first break down the problem statement, then discuss multiple approaches to solve the problem, and finally analyze the time and space complexities of these solutions.

## Problem Statement

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

```
Input: s1 = "bank", s2 = "kanb"
Output: True
Input: s1 = "attack", s2 = "defend"
Output: False
```

## Approach 1: Iterative Comparison

The most straightforward way to solve this problem is by iterating over the two strings character by character and keeping track of the indices where they differ.

### Algorithm Steps:

- Initialize an empty list called
`differences`

. - Loop through both strings character by character.
- If you find a pair of characters that don’t match, append the indices to
`differences`

. - After the loop, analyze the length of
`differences`

. Return`True`

or`False`

based on conditions.

#### Python Code:

```
def areAlmostEqual(s1: str, s2: str) -> bool:
differences = []
for i in range(len(s1)):
if s1[i] != s2[i]:
differences.append(i)
if len(differences) == 0:
return True
if len(differences) == 2:
i, j = differences
return s1[i] == s2[j] and s1[j] == s2[i]
return False
```

### Time and Space Complexity

- Time Complexity: O(n), where n is the length of the string.
- Space Complexity: O(k), where k is the number of differing characters between the two strings.

## Approach 2: List Comprehension

You can make the code more concise by employing list comprehensions to gather the indices of differing characters.

#### Python Code:

```
def areAlmostEqual(s1: str, s2: str) -> bool:
differences = [i for i, (c1, c2) in enumerate(zip(s1, s2)) if c1 != c2]
if len(differences) == 0:
return True
if len(differences) == 2:
i, j = differences
return s1[i] == s2[j] and s1[j] == s2[i]
return False
```

### Time and Space Complexity

- Time Complexity: O(n)
- Space Complexity: O(k)

## Conclusion

The ‘Check if One String Swap Can Make Strings Equal’ problem is an intriguing exercise in string manipulation and comparison. It provides various avenues for arriving at a solution, each presenting different Pythonic approaches and techniques.

Regardless of the approach used, the time complexity remains linear (O(n)O(n)), making all solutions efficient for this problem. The space complexity also remains linear in terms of differing characters.