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1085. Sum of Digits in the Minimum Number πŸ”’

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

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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

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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;
    }
}

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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;
    }
};

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

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