Sum of Digits in Base K

🏠 ⬅️ ➡️

Given an integer n (in base 10) and a base k, return the sum of the digits of n after converting n from base 10 to base k.

After converting, each digit should be interpreted as a base 10 number, and the sum should be returned in base 10.

Example 1:

Input: n = 34, k = 6 Output: 9 Explanation: 34 (base 10) expressed in base 6 is 54. 5 + 4 = 9.

Example 2:

Input: n = 10, k = 10 Output: 1 Explanation: n is already in base 10. 1 + 0 = 1.

Constraints:

  • 1 <= n <= 100
  • 2 <= k <= 10
Note: This problem is from LeetCode.
❌ Compiled
❌ Executed
❌ Correct 
program main
    implicit none

    integer :: n, k, result

    ! Example 1
    n = 34
    k = 6
    write (*, '(A, I0, A, I0, A, I0)') 'n = ', n, ', k = ', k, ', result = ', sum_digits(n, k)

    ! Example 2
    n = 10
    k = 10
    write (*, '(A, I0, A, I0, A, I0)') 'n = ', n, ', k = ', k, ', result = ', sum_digits(n, k)

contains

    function sum_digits(n, k) result(result)
        implicit none
        integer, value :: n, k
        integer :: result

        integer :: i, digit

        result = 0
        do i = 1, len(n)
            digit = modulo(n / (k ** (i - 1)), k)
            result = result + digit
        end do
    end function sum_digits
end program main
❌ Compiled
❌ Executed
❌ Correct 
```
program sum_digits
    implicit none
    integer :: n, k, i, sum
    character(len=10) :: num

    read(*,*) n, k

    ! Convert n from base 10 to base k
    num = ''
    do while (n > 0)
        num = char(ichar('0') + mod(n, k)) // num
        n = n / k
    end do

    ! Sum the digits of num in base 10
    sum = 0
    do i = 1, len_trim(num)
        sum = sum + ichar(num(i:i)) - ichar('0')
    end do

    ! Output the sum
    write(*,*) sum

end program sum_digits
```
🌐 Data from online sources 
def sum_base(n: int, k: int) -> int:
    s = 0
    while n:
        s += n % k
        n //= k
    return s

The algorithm is straightforward for converting a number from base 10 to base k and summing its digits. We use a while loop to repeatedly divide the number n by k, until n becomes 0. In each iteration, we add the remainder (n % k) to the sum, and update n by dividing it by k and taking the floor of the result. In the end, we return the sum.
🌐 Data from online sources 
int sumBase(int n, int k) {
    int sum = 0;
    while (n > 0) {
        sum += n % k;
        n /= k;
    }
    return sum;
}

The algorithm is straightforward for converting a number from base 10 to base k and summing its digits. We use a while loop to repeatedly divide the number n by k, until n becomes 0. In each iteration, we add the remainder (n % k) to the sum, and update n by dividing it by k and taking the floor of the result. In the end, we return the sum.