✅ Fastest and Easiest C++ Solution
Approach\nAs the size of both the strings is equal , iterate the strings and compare each consecutive characters and count the unequal characters . Since in the
# Approach\nAs the *size of both the strings is equal* , iterate the strings and compare each consecutive characters and *count the unequal characters* . Since in the *problem statement* it is clearly mentioned that we can swap 2 characters (*from exactly 1 string at a time*) for maximum 1 time. So the number of unequal characters should be **max 2 or 0**(in case the strings are already equal).\n\n\n\n**Note :** Here there might be some test cases where few characters are different in both the strings, I mean to say that there might be few characters which are present in s1 but not in s2, viceversa. In this case it is never possible to equalise the strings by a single swap , hence return false .\n\nExample : s1 = "aaz" , s2 = "caa"\n\n\n# Code\n```\nclass Solution {\npublic:\n    bool areAlmostEqual(string s1, string s2) {\n\n        //For checking the edge test cases , as mentioned above in Note\n        // copied the strings s1 and s2 in a and b respectively so that the original strings are not lost , and sorted strings a and b and compared them to find whether all characters in both the strings are same (maybe not in order)\n        string a = s1;\n        string b = s2;\n        sort(a.begin(),a.end());\n        sort(b.begin(),b.end());\n        if(a!=b) return false;\n       \n       //declared and initialised a counter variable to count the unequal characters from both the strings\n       int count = 0;\n\n       //traversed the strings of same length and incremented the counter variable in case unequal string characters are found\n       for(int i=0;i<s1.size();++i){\n           if(s1[i]!=s2[i]) count++;\n       } \n\n       /* 1.count 0 indicates that the strings are already equal && 2. count 2 indicates that with a single swap we can make the strings equal \n          In both the above cases we need to return true */\n       if(count==0 || count==2) return true;\n\n       return false;\n    }\n};\n```
['C++']