Let's delve into the coding intricacies of this concept:

Setting the Stage: The Base Case

Picture this: you're given a number, and your mission is to convert it into a binary string. But what if the number is less than 1? In this curious situation, there's no binary string to convert to! So, what do we do? Return an empty string, of course!

1if num < 1:
2    return ''

This empty string acts as a foundation for all further calculations.

The Case of the Odd One: num Not Divisible by 2

If your number isn't divisible by 2, it's the odd one out. And odd numbers have that pesky 1 hanging around at the end when converted to binary. How do we deal with it? By making it part of the process!

1if num % 2 > 0:
2    return decimalToBinary((num - 1) // 2) + '1'

A Practical Example with 3

If you encounter 3, which is an odd number, the above code snippet will effectively do this: 1. Subtract 1 to get 2. 2. Divide by 2 to get 1. 3. Convert 1 to binary, which is '1'. 4. Add the extra '1' from the odd number at the end.

Result: '11'

The Case of the Even Stevens: num Divisible by 2

If your number is perfectly divisible by 2, it's smooth sailing. Just halve the number and convert what's left into binary.

1if num % 2 == 0:
2    return decimalToBinary(num // 2) + '0'

These two conditions form the pillars of our decimal-to-binary conversion function. One handles the odd ones, adding an extra 1 to the binary string. The other tackles the even ones, adding a 0 as they go along.