Formatted question description: https://leetcode.ca/all/1085.html

# 1085. Sum of Digits in the Minimum Number

Easy

## Description

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. 1 <= A.length <= 100
2. 1 <= A[i].length <= 100

## Solution

Loop over the array A to find the minimum number. Then calculate the sum of the digits of the minimum number. Finally check whether the sum is odd or even, and return 0 or 1 respectively.

• class Solution {
public int sumOfDigits(int[] A) {
int min = Integer.MAX_VALUE;
for (int num : A)
min = Math.min(min, num);
if (min == 0)
return 1;
int sum = 0;
int temp = min;
while (temp > 0) {
sum += temp % 10;
temp /= 10;
}
return sum % 2 == 0 ? 1 : 0;
}
}

• // OJ: https://leetcode.com/problems/sum-of-digits-in-the-minimum-number/
// Time: O(N)
// Space: O(1)
class Solution {
public:
int sumOfDigits(vector<int>& A) {
int n = *min_element(begin(A), end(A)), sum = 0;
while (n) {
sum += n;
n /= 10;
}
return 1 - sum % 2;
}
};

• print("Todo!")