cpp solution || 100% faster || simplest approach
Thank you for checking out the solution\nDo upvote if it helps :)\n\n__Approach :\nFirst task is to check if the string is not a single character, if it is then
__Thank you for checking out the solution\nDo upvote if it helps :)__\n\n__Approach :\nFirst task is to check if the string is not a single character, if it is then no answer exists\nAfter this one important fact is to be used, "THE GIVEN STRING IS A PALINDROME"__\n```\nThis makes the takes simpler as any character at a distance x from the beginning \nis same as the character at a distance x from the end\nWhich means we only have to traverse half of the string\n```\n__So Start the traversal, if a character at the i(th) position is not \'a\'__\n>__Check if the length is odd and i = length/2__\n>> __If yes, break the loop (Since our string is like aaaa...*...aaaa), we h=will have to replace the last \'a\' with \'b\'__\n \n>__Change the character at the position i, to \'a\', this works because the character from the end is not same as \'a\' (NOW)\n>Return the manipulated String__\n>\n__If by now, nothing has been returned, the string is either all (a)s or of type  \naaaa...*...aaaa,  where * can be any character other than \'a\' and the length is surely odd for the string of this type__\n\n__What to do ?\nJust change the last \'a\' of the string to \'b\' and return the string__\n\n__Below is the coding implementation for the same__\n_____\n```\nclass Solution \n// Runtime: 0 ms, faster than 100.00% of C++ online submissions for Break a Palindrome.\n{\npublic:\n    string breakPalindrome(string palindrome)\n    {\n        ios_base::sync_with_stdio(0);\n        int l = palindrome.size();\n        // A single letter entry can\'t be broken in any way so that it doesnt stay a palindrome\n        // Hence return ""\n        if (l == 1)\n            return "";\n        // Now the remaining task to find a position that doesn\'t contain \'a\'\n        for (int i = 0; i <= l/2; i++)\n            if (palindrome[i] != \'a\')\n            {\n\t\t\t    //If the position lies in the center of the string, and the string has an odd-number length\n\t\t\t\t// Time to break, becoz changing it into \'a\', will create a new palindrome\n                if ((l & 1) && i == l / 2)\n                    break;\n\t\t\t\t// Otherwise change the character at index i to \'a\', and return it\n                palindrome[i] = \'a\';\n                return palindrome;\n            }\n        // In a case where\n        // We have all \'a\'s,\n        // Or when replacing a char in the middle makes the string palindrome again\n        // Just replace b with the character at the end\n        palindrome[l - 1] = \'b\';\n        return palindrome;\n    }\n};\n```\n__Time Complexity : O(N/2) ~ O(N) , where N is the length of the string\nSpace Complexity : O(1)__\n\n__Some Test case to consider:\n"aaaaaaa"\n"aaabaaa"\n"aaabbaaa"\n"abccba"\n"c"\n"a"__
['C++']