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.
 

Constraints:

1 <= A.length <= 100
1 <= A[i] <= 100











class Solution {
public:
    int sumOfDigits(vector<int>& A) {
        int m=INT_MAX;
        for(int i:A){
            m=min(m,i);
        }
        
        string n=to_string(m);
        int sum=0;
        
        for(char c:n){
            sum+=(c-'0');
        }
        return (sum+1)%2;
    }
};