BEST/ EASIEST APPROACH
Intuition   if size is 1 then empty because  every charcter is a palindrome . Check for half length if not found \'a\' then replace that character with \'a\' ot
# Intuition   if size is 1 then empty because  every charcter is a palindrome . Check for half length if not found \'a\' then replace that character with \'a\' otherwise replace the last character with \'b\'.\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach   check for a for making it lexicographically smallest if not found then replace the last character to \'b\' otherwise replace it with \'a\'.\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(n/2)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n    string breakPalindrome(string palindrome) {\n        int n=palindrome.size();\n        if(n==1){\n            return "";\n        }\n        bool ok=0;\n        for(int i=0;i<n/2;i++){\n            if(palindrome[i]!=\'a\'){\n                palindrome[i]=\'a\';\n                 ok=1;\n                break;\n            }\n        }\n        if(!ok)palindrome[n-1]=\'b\';\n        return palindrome;\n    }\n};\n```
['C++']