Clearest && Easiest Solution
Intuition\n Describe your first thoughts on how to solve this problem. \njust check if it is equal to the element at the same index or index + 2 and done\n\n# A
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\njust check if it is equal to the element at the same index or index + 2 and done\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n    bool canBeEqual(string s1, string s2) {\n        if (s1.size() != s2.size()) {\n        }\n        \n        for (int i = 0; i < s1.size(); i++) {\n            if (s1[i] == s2[i]) {\n            } else {\n                if (s1[i+2] == s2[i]) {\n                    swap(s1[i], s1[i+2]);\n                    }\n                }\n            }\n        \n        \n        return s1 == s2;\n    }\n};\n\n\n\n```
['C++']