Easiest O(N) Approach
Intuition1.Early Exit: Strings are already equal:2. Find the Mismatches:3. Check for Possible Swap:4. Return Result:Approach

Check if the strings are already e
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
1.Early Exit: Strings are already equal:

    If s1 and s2 are already the same, there’s nothing to swap. You can immediately return True. This case takes constant time and avoids unnecessary computations.

2. Find the Mismatches:

    If s1 and s2 are different, you need to find where the differences occur.
    You loop through the strings and compare corresponding characters. Each time you find a mismatch, you record the index.
    The key insight here is that if there are more than two mismatches, it's impossible to fix them with just one swap. This is because a swap can only fix two different positions, so if there are more mismatches, multiple swaps would be needed.
    If there are exactly two mismatches, there's still a chance that swapping can fix the strings, and you can proceed to the next step. 

3. Check for Possible Swap:

    After finding the mismatches, you need to check if a single swap can fix the problem.
    Specifically, if there are exactly two mismatched positions, the swap should satisfy:
    s1[i] == s2[j] and s1[j] == s2[i].
    This condition ensures that swapping the two characters in s1 at indices i and j would make the characters match their counterparts in s2.

4. Return Result:

    If there are no mismatches, the strings are already equal, so return True.
    If there are exactly two mismatches and the swap can fix the strings, return True.
    Otherwise, return False.
# Approach
<!-- Describe your approach to solving the problem. -->
1. Check if the strings are already equal:

    If s1 is equal to s2 from the start, return True. No swap is needed in this case.
    This is the simplest case and avoids unnecessary work. 

2. Find positions of mismatched characters:

    Traverse both strings character by character.
    Keep track of the indices where characters in s1 and s2 differ.
    If there are more than two mismatched positions, return False because you cannot fix more than two differences with a single swap.
    If there are exactly two mismatches, store the indices where the characters differ.

3. Check if the two mismatches can be swapped:

    If there are exactly two mismatched positions (let's call them i and j), check if swapping s1[i] with s1[j] results in s2[i] and s2[j] matching:
    Specifically, check if s1[i] == s2[j] and s1[j] == s2[i]. If this condition holds true, then a single swap can make the strings equal. 

4. Return the result:

    If there are no mismatches, return True because the strings are already equal.
    If there are exactly two mismatches and the swap condition holds, return True.
    Otherwise, return False.
# Complexity
- Time complexity: O(N)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->

- Space complexity: O(N)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->

# Code
```python3 []
class Solution:
    def areAlmostEqual(self, s1: str, s2: str) -> bool:
        if s1 == s2:
            return True
        e = []
        for i in range(len(s1)):
            if s1[i] != s2[i]:
                e.append(i)
        if len(e) == 2:
            i, j =e
            return s1[i] == s2[j] and s1[j] == s2[i]
        
        return False

```
['String', 'Counting', 'Python3']