Sherlock and the Beast – HackerRank

I found myself stuck on this problem recently. I must confess, I lost a couple of hours trying to get to figure the logic for this one. Here’s the problem:


I’ve written 2 functions to solve this problem. The first one I used for smaller N, say N < 30 and the second one for N > 30. The second function is elegant, and it relies on the mathematical property that if a number N is not divisible by 3, it could either leave a remainder 1 or 2.

If it leaves a remainder 2, then subtracting 5 once would make the number divisible by 3. If it leaves a remainder 1, then subtracting 5 twice would make the number divisible by 3.

We subtract 5 from N iteratively and attempt to divide N into 2 parts, one divisible by 3 and the other divisible by 5. We want the part that is divisible by 3 to be the larger part, so that the associated Decent Number is the largest possible. This explanation might seem obtuse, but if you get pen down on paper, you’ll understand what I mean.