class Solution:
    '''
    Time Complexity: O(l)
    Space Complexity: O(l)
    where, l is number of digits in n
    Solved in 7 mins, after 1 hint! Ok job!
    '''
    def minPartitions(self, n: str) -> int:
        if '9' in n: # Since we can only use digits 0 and 1 in deci-binary numbers, if there is a digit '9' in n, we will need at least 9 deci-binary numbers to sum up to it, so we can directly return 9 in that case, otherwise we need to find the maximum digit in n, which will be the minimum number of deci-binary numbers needed to sum up to n, since each deci-binary number can contribute at most 1 to each digit place.
            return 9
        return int(max(c for c in n))