Sum of Digits in the Minimum Number

🏠 ⬅️ ➡️

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
Note: This problem is from LeetCode.
❌ Compiled
❌ Executed
❌ Correct 
program main
    implicit none

    integer, parameter :: n = 8
    integer, dimension(n) :: nums = (/ 34, 23, 1, 24, 75, 33, 54, 8 /)
    integer :: i, min_index, sum_digits, result

    ! find the index of the minimum element in nums
    min_index = minloc(nums, dim=1)

    ! calculate the sum of the digits of the minimum element
    sum_digits = 0
    do i = 1, len_trim(nums(min_index))
        sum_digits = sum_digits + mod(nums(min_index), 10)
        nums(min_index) = nums(min_index) / 10
    end do

    ! return 0 if the sum of the digits is odd, otherwise return 1
    result = merge(0, 1, mod(sum_digits, 2) == 0)

    print '(I0)', result

end program main
✅ Compiled
✅ Executed
⚠️ Correct 
module min_digit_sum
  implicit none
contains
  function solve(nums) result(output)
    integer, intent(in) :: nums(:)
    integer :: output
    integer :: min_num, sum_digits

    ! Find the minimum element in nums
    min_num = minval(nums)

    ! Calculate the sum of the digits of the minimum element
    sum_digits = 0
    do while (min_num > 0)
      sum_digits = sum_digits + mod(min_num, 10)
      min_num = min_num / 10
    end do

    ! Return 0 if the sum of the digits is odd, or 1 otherwise
    output = 1
    if (mod(sum_digits, 2) == 1) then
      output = 0
    end if
  end function solve
end module

program main
  use min_digit_sum
  implicit none
  integer, parameter :: nums(8) = [34, 23, 1, 24, 75, 33, 54, 8]
  integer :: output

  output = solve(nums)
  write (*, '(I1)') output
end program
🌐 Data from online sources 
def sum_of_digits(nums):
    min_val = min(nums)
    digit_sum = sum(int(digit) for digit in str(min_val))
    return 1 if digit_sum % 2 == 0 else 0

1. Find the minimal element (`minVal`) of given array `nums`.
  1. Calculate the sum of the digits of the minimal element (digitSum).
  2. Check if the sum of the digits is odd or even.
  3. If it's even, return 1.
  4. If it's odd, return 0.
🌐 Data from online sources 
int sumOfDigits(vector<int>& nums) {
    int minVal = *min_element(nums.begin(), nums.end());
    int digitSum = 0;
    while (minVal > 0) {
        digitSum += minVal % 10;
        minVal /= 10;
    }
    return digitSum % 2 == 0 ? 1 : 0;
}

1. Find the minimal element (`minVal`) of given array `nums`.
  1. Calculate the sum of the digits of the minimal element (digitSum).
  2. Check if the sum of the digits is odd or even.
  3. If it's even, return 1.
  4. If it's odd, return 0.