Boundary case for Delhi's Odd Even coding problem

#1

According to the Description of the problem, we need to print “Yes”, if the sum of digits which are even is divisible by 4 or sum of digits which are odd in that number is divisible by 3.

If we take the input as 1, then Sum of odd digits is 1 and there cannot be sum of even digits as there are no even digits.
As 1 is not divisible by 3, It should print “No”. But the solution is not getting accepted for this.

Code link : https://ide.codingblocks.com/s/417166

If we consider the sum of even digits as 0 for the same input, then answer will be “Yes” as 0 is divisible by 4 and this solution is being accepted.

Code link: https://ide.codingblocks.com/s/417163

In this case the answer will always be “Yes” if there are no Odd digits or if there are no Even digits.
Ex: 11111111, 22222

I think that we shouldn’t consider the sum of even numbers if there are no even numbers in the Car number and similarly for odd numbers.
Please clarify.

Also, If we unlock the test cases after we scored max marks for the question, will those marks not considered?
My understanding is that we can’t score more marks than what we already scored. But it is mentioned that I will not get any marks if I opened the test cases.
Please clarify.

#2

Your score will not be updated by your submissions after you open the test cases or unlock the tutorial.

#3

The Question in Simple language is that:

  • You have given a number and in that number, the digits which are even should have sum divisible by 4 or the sum of digits which are odd should be divisible by the 3, Only then you can say yes otherwise you will say no.

Algorithm

  • Take input of the Number.
  • Declare two variables to store the sum of even numbers and odd numbers.
  • Extract digit one by one ( by % 10).

    1. Check if the number is even or odd.

    2. If even add the number in the variable storing even sum.

  1. Othewise add it in the variable storing odd sum.
  • After the loop, Check if the even sum is divisible by 4.

    1. If True, print Yes.

    2. otherwise print No.

  • Check if the odd sum is divisible by 3.

    1. If True, print Yes.

    2. otherwise print No.