Simple, beats 100% t, 85.22%s
Intuition\n\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Tim
# Intuition\n![Screenshot 2024-11-17 224142.png](https://assets.leetcode.com/users/images/bc45a910-e3e7-43ba-bf71-252148cd2d6c_1731879748.6760402.png)\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: $$O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```python3 []\nclass Solution:\n    def canBeEqual(self, s1: str, s2: str) -> bool:\n        list_s1= list(s1)\n        list_s2= list(s2)\n        \n        for i in range(0, len(s1)-2):\n            if list_s1[i]!=list_s2[i]:\n                swap = list_s1[i]\n                list_s1[i] = list_s1[i+2]\n                list_s1[i+2] = swap\n            if list_s1==list_s2:\n                return True\n        return False\n\n\n\n    \n            \n            \n\n\n\n\n\n        \n        \n\n\n\n\n\n        \n```
['Python3']