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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 <= 1001 <= nums[i] <= 100
Solutions
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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; } } -
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 }