Convert Integer to the Sum of Two No-Zero Integers

🏠 ⬅️ ➡️

No-Zero integer is a positive integer that does not contain any 0 in its decimal representation.

Given an integer n, return a list of two integers [a, b] where:

  • a and b are No-Zero integers.
  • a + b = n

The test cases are generated so that there is at least one valid solution. If there are many valid solutions, you can return any of them.

Example 1:

Input: n = 2 Output: [1,1] Explanation: Let a = 1 and b = 1. Both a and b are no-zero integers, and a + b = 2 = n.

Example 2:

Input: n = 11 Output: [2,9] Explanation: Let a = 2 and b = 9. Both a and b are no-zero integers, and a + b = 9 = n. Note that there are other valid answers as [8, 3] that can be accepted.

Constraints:

  • 2 <= n <= 104
Note: This problem is from LeetCode.
✅ Compiled
✅ Executed
⚠️ Correct 
program main
    implicit none

    integer :: n, a, b

    n = 2
    call solve(n, a, b)
    print '(I0, " ", I0)', a, b

    n = 11
    call solve(n, a, b)
    print '(I0, " ", I0)', a, b

contains

    subroutine solve(n, a, b)
        implicit none
        integer, intent(in) :: n
        integer, intent(out) :: a, b

        if (mod(n, 2) == 0) then
            a = n / 2
            b = n - a
        else
            a = (n - 1) / 2
            b = n - a
        end if
    end subroutine solve
end program main
✅ Compiled
✅ Executed
⚠️ Correct 
module no_zero_integers

implicit none

contains

function solve(n) result(a_b)

integer, intent(in) :: n
integer :: a, b, a_b(2)

a = 1
b = n - a

if (a == 0 .or. b == 0) then
    a = 1
    b = n - a
end if

a_b(1) = a
a_b(2) = b

end function solve

end module no_zero_integers

program main

use no_zero_integers

implicit none

integer :: n
integer :: a_b(2)

n = 2
a_b = solve(n)

write (*,*) a_b

n = 11
a_b = solve(n)

write (*,*) a_b

end program main
🌐 Data from online sources 
def no_zero_integers(n: int):
    for a in range(1, n):
        if '0' not in str(a) and '0' not in str(n - a):
            return [a, n - a]
    return [0, 0]  # never gets here

We will write a function that receives an integer n as the input. The algorithm then iterates over all integers from 1 to n-1 as a candidate for value a in the result [a, b]. For each candidate a, we check if both a and n-a (which would be b) are No-Zero integers by ensuring that neither of the strings representation have a '0'. If both strings don't have a '0', we found a valid answer and return a list containing [a, n-a]. The loop should always find a valid answer for the given constraints.

🌐 Data from online sources 
std::vector<int> noZeroIntegers(int n) {
    for (int a = 1; a < n; ++a) {
        if (std::to_string(a).find('0') == std::string::npos &&
            std::to_string(n - a).find('0') == std::string::npos) {
            return {a, n - a};
        }
    }
    return {0, 0}; // never gets here
}

We will write a function that receives an integer n as the input. The algorithm then iterates over all integers from 1 to n-1 as a candidate for value a in the result [a, b]. For each candidate a, we check if both a and n-a (which would be b) are No-Zero integers by ensuring that neither of the strings representation have a '0'. If both strings don't have a '0', we found a valid answer and return a list containing [a, n-a]. The loop should always find a valid answer for the given constraints.