easy solution
Intuitionbrute force through both stringsApproachComplexity
Time complexity:

Space complexity:
Code
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
brute force through both strings

# Approach
<!-- Describe your approach to solving the problem. -->

# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->

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

# Code
```python3 []
class Solution:
    def canBeEqual(self, s1: str, s2: str) -> bool:
        first = list(s1)
        second = list(s2)
        origf, origs = first.copy(), second.copy()
        for i in range(len(first)):
            for j in range(len(first)):
                if abs(j - i) == 2:
                    print(i, j)
                    first[i], first[j] = first[j], first[i]
                    print(first)
                    if first == second:
                        return True  
        for i in range(len(second)):
            for j in range(len(second)):
                if abs(j - i) == 2:
                    second[i], second[j] = second[j], second[i]
                    if second == origf:
                        return True  
                    second[:] = origs
        return False
```
['Python3']