Sum of digits in minimum number

Time: O(NxL); Space: O(l); easy

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: A = [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: A = [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.

Notes:

  • 1 <= len(A) <= 100

  • 1 <= len(A[i]) <= 100

[2]:

class Solution1(object):
    def sumOfDigits(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        total = sum([int(c) for c in str(min(A))])
        return 1 if total % 2 == 0 else 0

[3]:

s = Solution1()
A = [34,23,1,24,75,33,54,8]
assert s.sumOfDigits(A) == 0
A = [99,77,33,66,55]
assert s.sumOfDigits(A) == 1