✔ || C++ Easy Solution with edge case Explanation backtrack
\nclass Solution {\npublic:\n    bool ispal(string s){\n        int l=0,r=s.size()-1;\n        while(l<r)\n        if(s[l++]!=s[r--]) return false;\n        ret
```\nclass Solution {\npublic:\n    bool ispal(string s){\n        int l=0,r=s.size()-1;\n        while(l<r)\n        if(s[l++]!=s[r--]) return false;\n        return true;\n    }\n    string breakPalindrome(string pal) {\n        if(pal.size()==1) return "";\n        for(int i=0;i<pal.size();i++){\n            if(pal[i]!=\'a\'){\n                char ch=pal[i];\n                pal[i]=\'a\';\n                if(ispal(pal)) {pal[i]=ch; break;} // Lets assume "aba" string, our string become "aaa" it is palindrome ,so here we have to cheak after operation string become palindrome or not , if palindrome then make the string back to previous one and change the last element to \'b\n                return pal;\n            }\n        }\n        pal[pal.size()-1]=\'b\';\n        return pal;\n    }\n};\n**If U Like It Please Upvote.**\n```
['Greedy', 'C']