Question

1085. Sum of Digits in the Minimum Number

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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

Solution 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

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;
  }
};

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