# [Haskell-cafe] For Project Euler #1 isn't it more efficient to generate just the numbers you need? <Spoiler>

KC kc1956 at gmail.com
Sat May 7 20:52:51 CEST 2011

```Do you mean O(1) complexity?

Don't you have to touch at least the multiples of 3 & 5 making it O(k)
where is the number of multiples of 3 and the number of multiples of
5?

On Fri, May 6, 2011 at 10:10 PM, Lyndon Maydwell <maydwell at gmail.com> wrote:
> If you're looking for efficiency, I believe you can actually do #1 in
> constant time:
>
>
> On Sat, May 7, 2011 at 7:31 AM,  <caseyh at istar.ca> wrote:
>> -- sumMultiples3or5 s = sum [x | x <- [3..s-1], x `mod` 3 == 0 || x `mod` 5
>> == 0]
>>
>>
>> -- Isn't this faster
>>
>> sumMultiples3or5 s = sum ([x | x <- [3,6..s-1]] `merge` [x | x <-
>> [5,10..s-1]])
>>
>> merge xs [] = xs
>> merge [] ys = ys
>> merge txs@(x:xs) tys@(y:ys)
>>    | x < y     = x : xs `merge` tys
>>    | x > y     = y : txs `merge` ys
>>    | otherwise = x : xs `merge` ys
>>
>>
>>
>> _______________________________________________
>>
>
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