Let us take a breather by tackling a problem simple enough that its solution needs only a couple of conditional statements and some arithmetic, but not even one loop or anything more fancy. The difficulty is coming up with the conditions that cover all possible cases of this problem just right, including all of the potentially tricksy edge and corner cases, without being off-by-one anywhere.
Given three integers
start, end, n, so that
start <= end, count how many integers between
end, inclusive, are evenly divisible by
n. Note that either
end can well be negative or zero, but
n is guaranteed to be greater than zero. Here is the scheme for
1, 9, 3 input.
Sure, this problem could be solved as a one-liner with the lazy comprehension. Try to have code with no loops at all, but use only integer arithmetic and conditional statements to root out the truth.
Input: Three integers (int).
Output: Integer (int).
assert count_divisibles(7, 28, 4) == 6 assert count_divisibles(-77, 19, 10) == 9 assert count_divisibles(1, 999999999999, 5) == 199999999999
- start <= end;
- n > 0.
The mission was taken from Python CCPS 109. It is taught for Ryerson Chang School of Continuing Education by Ilkka Kokkarinen
CheckiO Extensions allow you to use local files to solve missions. More info in a blog post.
In order to install CheckiO client you'll need installed Python (version at least 3.8)
Install CheckiO Client first:
pip3 install checkio_client
Configure your tool
checkio --domain=py config --key=
Sync solutions into your local folder
(in beta testing) Launch local server so your browser can use it and sync solution between local file end extension on the fly. (doesn't work for safari)
checkio serv -d
Alternatevly, you can install Chrome extension or FF addon
checkio install-plugin --ff
checkio install-plugin --chromium