beats 99% simple 4 conditions
Intuition\n Describe your first thoughts on how to solve this problem. \nThe goal is to transform a given palindrome string into a non-palindrome string that is
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe goal is to transform a given palindrome string into a non-palindrome string that is the smallest possible in lexicographical order. Here\'s how to approach it step-by-step.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Edge Case: If the length of the palindrome is 1, return an empty string since it\'s impossible to break the palindrome.\n2. Replace First Non-\'a\' Character: Iterate through the string. For the first character that is not \'a\', replace it with \'a\'. If this transformation makes the string non-palindromic, return it immediately.\n3. Check if Result is Still a Palindrome: After replacing the first non-\'a\' character, check if the resulting string is a palindrome. If it is, continue to the next character.\n4. All Characters are \'a\': If the entire string consists of \'a\'s and replacing the first non-\'a\' character still results in a palindrome, replace the last character with \'b\' to ensure the string is non-palindromic.\n\n# Complexity\n- Time complexity: O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(n)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n    def breakPalindrome(self, palindrome: str) -> str:\n        if len(palindrome)==1:\n            return ""\n        else:\n            for x in range(len(palindrome)):\n                s = ""\n                if palindrome[x]!="a":\n                    s+=palindrome[:x]+\'a\'+palindrome[x+1:]\n                    if s!=s[::-1]:\n                        return s\n            s = palindrome[:len(palindrome)-1]+"b"\n            return s\n```
['String', 'Greedy', 'Python3']