Easy to Understand || 100% faster || O( n ) || C++
\nclass Solution {\npublic:\n    string breakPalindrome(string pal){\n        int n = pal.size();\n        if(n == 1){\n            return "";\n        }\n\t\t\
```\nclass Solution {\npublic:\n    string breakPalindrome(string pal){\n        int n = pal.size();\n        if(n == 1){\n            return "";\n        }\n\t\t\n\t\t// traverse only half of the string\n\t\t// if we find any ith character other than \'a\' make that ith character to \'a\' \n\t\t// return the string\n\t\t\n        for(int i=0; i<n/2; i++){\n            if(pal[i] != \'a\'){\n                pal[i] = \'a\';\n                return pal;\n            }\n        }\n\t\t\n        // if all character in string are only \'a\'.  \n\t\t// Ex : "aaaaa" then make the last character to \'b\' -> "aaaab"\n\t\t\n\t\tpal[n-1] = \'b\'; \n        return pal;\n    }\n};\n```\n: ) \uD83D\uDC4D : )
['String', 'C', 'C++']