C#•2y ago

❔ Efficient way to query this problem

I have a logic question. Right now, I'm doing it the worst way possible... Which is s adly the only way I know rn. Nesting for loops. I know, I know. Hence why I'm looking for a better way

33 Replies

Angius•2y ago

You can use LINQ It won't be as performant, but will be more concise than nested loops SelectMany() will be of particular interest

sentrycodOP•2y ago

Hmm, how would linq go about this if I may ask

Angius•2y ago

var allInvestigations = someCase.Visits
  .SelectMany(visit => visit.Investigations)
  .ToArray();

var allInvestigations = someCase.Visits
  .SelectMany(visit => visit.Investigations)
  .ToArray();

sentrycodOP•2y ago

I'd give this a shot thanks. You said it won't be as performant. Do you mean it'd be less performant than the for loop method?

Anton•2y ago

what's wrong with nesting for loops

sentrycodOP•2y ago

Bad time complexity

Angius•2y ago

Yes, it will most probably be less performant Whether the difference matters or not, up to you

Anton•2y ago

huh? what's the time complexity, you figure? name it

sentrycodOP•2y ago

The larger the list grows, the more time it'd take to query

Angius•2y ago

That will be true regardless of how you handle it

Anton•2y ago

there are two lists

Relevant•2y ago

You're getting all elements regardless

Anton•2y ago

no, let them think it through you have no right to worry about time complexity unless you understand it

sentrycodOP•2y ago

So you're saying the time to carry this out won't change regardless of method used

Anton•2y ago

I urge you to think it through if you want to do some operation on all pairs (visit, investigation), you'll have to go through all pairs how many pairs are there?

sentrycodOP•2y ago

Just two

Relevant•2y ago

There are certainly ways to make it really bad if you tried hard enough. So I wouldn't say regardless of method used, but there will be a limit to how few steps is possible

Anton•2y ago

the number of pairs is the same as the optimal time complexity think again if there are two visits and three investigations, how many pairs are there?

sentrycodOP•2y ago

2 to the 3rd power?

Anton•2y ago

(a,b) and (1,2,3) how many pairs (a,1), (a,2), ... list them ok let me List them (a,1), (a,2), (a,3), (b,1), (b,2), (b,3) are these all the pairs?

sentrycodOP•2y ago

Given 2 visits and 3 investigations I guess so

Anton•2y ago

see any patterns that might lead you to a formula? a formula for the number of pairs

sentrycodOP•2y ago

3n? Where n is the number of visits?

Anton•2y ago

how many choices do you have for the first element? how many choices for the second element for each of the choices for the first element? yeah, because you have n choices for the first element and 3 choices for the second element for each of the first choices 3 is what the size of what?

sentrycodOP•2y ago

Number of objects in investigation list?

Anton•2y ago

yeah call it m so what's the number of pairs, in terms of n and m?

sentrycodOP•2y ago

3mn?

Anton•2y ago

think again? Now think about what a nested for loop does. It goes through all visits (the choice for the first element) and for each of those goes through each investigation (the second choice for each of the first choices). If you think about it for a bit, you will realize it's fundamentally same number of iterations as the number of pairs. you could do an unnested loop through all pairs instead. ig like this

foreach (var (v, i) in visits.SelectMany(v => investigations.Select(i => (v, i))))
    // operation

foreach (var (v, i) in visits.SelectMany(v => investigations.Select(i => (v, i))))
    // operation

is this less operations than a nested for loop? oh i reread the question just now

sentrycodOP•2y ago

Thanks in advance? Allow me to go learn more on this to fully get what you mean here

Anton•2y ago

sure in your question you can't possibly do it any other way, since the investigations are inside the visits you'll have to go through all visits, whatever you do there's fundamentally no way around it

sentrycodOP•2y ago

Ah I see. Damn. Thanks for taking your time with this man

Angius•2y ago

Man, I'd just stop at "unga bunga linq" there lol

Accord•2y ago

Was this issue resolved? If so, run /close - otherwise I will mark this as stale and this post will be archived until there is new activity.

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❔ Efficient way to query this problem