Saturday, August 6, 2011

Project Euler Problem #1

Alright, here we go on Project Euler. Time for the first problem:

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.

That doesn't sound quite that hard. Right? The first problem couldn't possibly be hard.

Well let's break it down. All the numbers that are multiples of 3 or 5 under 10 are 3, 5, 6 and 9. Or in other words, 3, 5, 6 and 9 are the only numbers (under 10) that are divisible by 3 or 5 and leave no remainder behind.

So now we have to find all the numbers under 1000 that satisfy the same requirement. And then sum those numbers.

At first thought I suppose it would be easy enough to create a loop that goes through each possibility and adds the number to a list if it passes the divisible test.

Something like this:

 List<int> multiples = new List<int>();

 for (int i = 1; i < 1000; i++)
  if ((i % 3 == 0) || (i % 5 == 0))

 return multiples.Sum();

Simple enough right? Well yes and for this simple type of problem, it won't really have any affect on performance. However, I wanted the least amount of lines as possible. I had recently discovered the wonderful world of lambda expressions and I feel that I should use them as often as I can. (for better or for worse)

So here is my solution:

 return Enumerable.Range(1, 999).Where(i => (i % 3).Equals(0) || (i % 5).Equals(0)).Sum();

Ah. Enumerables. By using the 'Range' extension I can retrieve an enumerable list of ints to select from. The 'Where' extension method will ensure that I only retrieve the numbers that are multiples of 3 or 5. And finally, the 'Sum' extension will return the sum of all the numbers left.

Or if we want to get really fancy, we could create a lambda function delegate to make things a bit more readable:

 Func<int, int, bool> IsMultiple = (int i, int j) => (i % j).Equals(0);
 return Enumerable.Range(1, 999).Where(i => IsMultiple(i, 3) || IsMultiple(i, 5)).Sum();

However, the previous solution will do in this case.

Simple and concise. KISS principle in action.

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