Simple || Easy to understand || Using sorting || C++
Intuition\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 
# Intuition\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:\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        string x=s1;\n        string y=s2;\n        \n        sort(x.begin() , x.end());\n        sort(y.begin() , y.end());\n        \n        if(x!=y)\n        return 0;\n        \n        for(int i=0 ; i<=1 ; i++)\n        {\n            if(s1[i]==s2[i])\n            continue;\n            else\n            {\n                if(s1[i]==s2[i+2])\n                {\n                    swap(s1[i] , s1[i+2]);\n                }\n                \n            }\n            \n            if(s1==s2)\n            return 1;\n        }\n        \n        if(s1==s2)\n        return 1;\n        \n        return 0;\n    }\n};\n```
['C++']