1085. Sum of Digits in the Minimum Number

Description

Given an integer array nums, return 0 if the sum of the digits of the minimum integer in nums is odd, or 1 otherwise.

Example 1:

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


Example 2:

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


Constraints:

• 1 <= nums.length <= 100
• 1 <= nums[i] <= 100

Solutions

• class Solution {
public int sumOfDigits(int[] nums) {
int x = 100;
for (int v : nums) {
x = Math.min(x, v);
}
int s = 0;
for (; x > 0; x /= 10) {
s += x % 10;
}
return s & 1 ^ 1;
}
}

• class Solution {
public:
int sumOfDigits(vector<int>& nums) {
int x = *min_element(nums.begin(), nums.end());
int s = 0;
for (; x > 0; x /= 10) {
s += x % 10;
}
return s & 1 ^ 1;
}
};

• class Solution:
def sumOfDigits(self, nums: List[int]) -> int:
x = min(nums)
s = 0
while x:
s += x % 10
x //= 10
return s & 1 ^ 1


• func sumOfDigits(nums []int) int {
s := 0
for x := slices.Min(nums); x > 0; x /= 10 {
s += x % 10
}
return s&1 ^ 1
}