Next palindrome using same digits

We will find the next permutation that is greater than the first half of the palindrome, say greaterHalf. Then add it to the mid item if there is one and to reversed of greaterHalf.

class Solution:
    def nextPalindrome(self, num: str) -> str:
        digits = list(num)
        mid = len(num) // 2
        midStr = "" if (len(num) % 2 == 0) else digits[mid]

        leftGreater = self.getNextLarger(digits[:mid])
        if not leftGreater:
            return ""

        return  "".join(leftGreater) + midStr + "".join(reversed(leftGreater))

    def getNextLarger(self, digits: List[str]) -> Optional[List[str]]:
        i = len(digits) - 2
        while i >= 0 and digits[i] >= digits[i+1]:
            i -= 1
        if i < 0:
            return []

        j = len(digits) - 1
        while j>=0 and digits[i] >= digits[j]:
            j -= 1

        digits[i], digits[j] = digits[j], digits[i]
        digits[i+1:] = reversed(digits[i+1:])

        return digits

Time complexity: O(n)
Space complexity: O(n)