Python3 - O(n / 2) Time | O(n) Space | With Comments.
\nclass Solution:\n\n# intuition is simple\n# the moment we encounter a character other than an \'a\', replace it with \'a\' (to make it lexographically shortes
```\nclass Solution:\n\n# intuition is simple\n# the moment we encounter a character other than an \'a\', replace it with \'a\' (to make it lexographically shortest)\n# if we do not find any other character, it means it\'s all filled with \'a\'s\n# so the next shortest string then comes when we replace the last character with a \'b\'\n\n    def breakPalindrome(self, palindrome: str) -> str:\n        n = len(palindrome)\n        if n == 1:\n            return ""\n        \n\t\t# \'flag\' to check if we encounter an \'a\'\n\t\t# also, we want to iterate through half the string, leaving the middle element if length is odd\n\t\t# in each iteration, we fill the resultant array from both the ends\n\t\t# rest is intuitive\n\t\t\n        res, flag = [palindrome[n // 2]] * n, 0\n        for i in range(n // 2):\n            if not flag and palindrome[i] != \'a\':\n                res[i] = \'a\'\n                flag = 1\n            else:\n                res[i] = palindrome[i]\n            res[n - i - 1] = palindrome[i]\n            \n        if not flag:\n            res[-1] = "b"\n        return "".join(res)\n```
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