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.
Note:
1 <= A.length <= 100 1 <= A[i].length <= 100
// 4ms, 59%
class Solution {
public:
int sumOfDigits(vector<int>& A) {
if (A.empty()) return -1;
int v = *min_element(A.begin(), A.end());
int r = 0;
while (v > 0) {
r += v % 10;
v /= 10;
}
return (r % 2) == 0;
}
};