 # Hackerrank optimizing euler problem #1

Hello everyone,

I spent my weekend optimising the first problem and learned allot in the process about Elixir. I got to 80 points, but still the third hidden test is timing out. Can anyone take a look at the code bellow and give me some pointers on how to improve the performance of the algorithm?

Problem is here

You can find the solution bellow; notice that I tried to parallelise as much as I could where it made sense (i.e. the solution is being computed as you input the numbers, in parallel). I don’t have allot of experience with Elixir so I would appreciate a helping hand.

``````defmodule Solution do

def calculateFinalSum([{:ok, a}, {:ok, b}, {:ok, c}]) do
a+b-c
end

def dividedSum(n, dividend) do
p = div(n-1, dividend)
div(dividend * p * (p+1), 2)
end

def solveOne(n) do
{number, _} = Integer.parse(n)
|> Enum.to_list
|> Solution.calculateFinalSum
end

def solve() do
IO.gets("")
IO.stream(:stdio, :line)
|> Enum.each(fn {:ok, n} -> IO.puts(n) end)
end
end

Solution.solve
`````` Tasks seem unnecessary here. They probably cost more than the value they add over sequentially calling `dividedSum/2`.

2 Likes

Most Project Euler questions revolve around mathematical insight. This particular question has an O(1) solution!

A hint to point you in the right direction is to consider “triangle numbers”, or the 3rd line of Pascal’s triangle. The sum of all the numbers between 0 and 100 can be thought of as (0 + 100) + (1 + 99) + (2 + 98) + (3 + 97) … + (49 + 51) + 50. Using this sort of procedure, you can find the sum of any sequence of whole numbers from 0 to n by this equation:

(n^2)/2 + n/2, which is the same as n(n + 1) / 2

Can you think of a way to find the sum of all the numbers from 0 to n that are divisible by 3? Or the numbers divisible by 5? Or 15?

2 Likes

Hello,

@AlchemistCamp, This is exactly what I am using here, its a bit hidden by the optimizations I tried to make. If you read carefully, I do: 3*(1+2+…+n/3)+5*(1+2+…+n/5)-15*(1+2+…+n/15), where 1+2+…+n/3 and the other series are calculated using the formula Sn = n*(n+1)/2.

@idiot, Indeed, you were correct. Getting rid of the tasks solved my issues. This was the winning solution:

``````defmodule Solution do

def dividedSum(n, dividend) do
p = div(n-1, dividend)
div(dividend * p * (p+1), 2)
end

def solveOne(n) do
Solution.dividedSum(n, 3)+Solution.dividedSum(n, 5)-Solution.dividedSum(n, 15)
end

def solve() do
{lines, _ }= IO.gets("") |> Integer.parse
for _ <- 1..lines do
{number, _} = IO.gets("") |> Integer.parse
number
|> Solution.solveOne
|> IO.puts
end
end
end

Solution.solve

``````

Thanks for the help! In the end, I overcomplicated things.

1 Like

So I see! I was too quick to write that after seeing the `Enum.each` in `solve`.

Is the current code still not passing the last test?

Edit: My carelessness was the root cause, but maybe renaming `dividedSum` to something like `sum_of_multiples(n, multiple)` would make it clearer, also.

The current code, now passes all tests 2 Likes