BREAK A PALINDROME ||  EASY TO UNDERSTAND || WELL COMMENTED || SIMPLE SOLUTION
\n/*\nfor making it lexographically smaller non-palindromic after replacing a character , we will simply look for a character which is not \'a\' from i=0 to i<n
```\n/*\nfor making it lexographically smaller non-palindromic after replacing a character , we will simply look for a character which is not \'a\' from i=0 to i<n/2 .\n1.If such character exists , we will simply put palindrome[i]=\'a\' and exists out of the loop while making flag=1 (there is such character exists)\n2. If such character doesn\'t exist then flag will remain equal to 0 .In this case , we will simply put palinrome[n-1]=\'a\';\n*/\nclass Solution {\npublic:\n    string breakPalindrome(string palindrome) {\n        int  n=palindrome.size();//calculating size\n//there is no sense of performing if size<2 as there is no way to make it non-palindrome\n        if(n<2)\n        {\n            return "";\n        }\n    int flag=0;\n        \n        for(int i=0;i<n/2;i++)\n        {\n            if(palindrome[i]!=\'a\')\n            {\n                flag=1;\n                palindrome[i]=\'a\';\n                break;\n            }\n        }\n        if(flag==0)\n        {\n            palindrome[n-1]=\'b\';\n        }\n        return palindrome;\n    }\n};\n```\n**IF YOU FOUND THIS HELPFUL , PLEASE DO UPVOTE IT**
['String', 'C']