"""
Given an array A of positive integers, let S be the sum of the digits of the minimal element of A.

Return 0 if S is odd, otherwise return 1.

Example 1:

Input: [34,23,1,24,75,33,54,8]
Output: 0
Explanation: 
The minimal element is 1, and the sum of those digits is S = 1 which is odd, so the answer is 0.


Example 2:
Input: [99,77,33,66,55]
Output: 1
Explanation: 
The minimal element is 33, and the sum of those digits is S = 3 + 3 = 6 which is even, so the answer is 1.
"""

class Solution:
    def sumOfDigits(self, A: List[int]) -> int:
        minNum = min(A)
        n = 0
        while minNum > 0:
            n += minNum % 10
            minNum //= 10
        return 0 if n%2 else 1
