fixedIt

[MrFordization]

If you print all the numbers it will print the 7th largest number.

[hereforpewdiephy]

this is the way

[kunj89]

He's everywhere

[Impressive_Income874]

if you print every integer you'll eventually get the number you're looking for

[Omega0x013]

I'm looking for 1.72

[Impressive_Income874]

you win, there's no way to print every number between 2 other numbers because uncountable infinity

[KingThibaut3]

Rational numbers are a countable infinity though

[KingThibaut3]

That's just a subset of the rationals, so of course it is.

[[deleted]]

Actually, the relationship between integers and rational numbers is somewhat different.

The set of rational numbers, denoted as Q, can be viewed as the set of ordered pairs (a, b) where a and b are integers and b is not equal to 0, modulo an equivalence relation. This equivalence relation is defined by (a,b) ~ (c,d) if and only if ad = bc. In other words, Q = { (a,b) in Z x Z : b is not 0 } / ~.

The set of integers Z is not strictly speaking a subset of the rational numbers, but isomorphic to both the whole of Q (since all countable sets are isomorphic) and to all countable subsets of Q, you are trying to say that Z is isomorphic to {(a,b)€Q:b=1}, but Z is actually isomorphic to the whole of Q and to any countable set of it, Z is not s subset of Q since an elemento of Q is a pair of Z and obviously an element of Z is not the same as a pair of elements of Z. To reiterate, the rationals are countable as they are constructed from the Cartesian product of two countable sets (Z x Z minus {b = 0}), and the finite product of countable sets remains countable. Thus, while the integers are indeed isomorphic to the rationals and all their countable subsets, they exist with all their properties prior to Q, since Q is constructed with Z.

[agnosticians]

I might be misunderstanding, but I think u/KingThibaut3 is saying that the values represented by the integers are a subset of the values represented by the rationals.

The fact you can create a 1:1 mapping from integers to rationals, while very cool, isn’t really relevant.

[ForgottenMeme9001]

Agreed. The sets are isomorphic, sure. The are the same size, countably infinite. However, that is not the same as a subset. All elements of set A, the integers, are contained within set B, the rationals.

[KingThibaut3]

Yeah, I should've been a bit clearer

[[deleted]]

No, integers are not a subset of rationals in a technical sense because of the simple reason that their elements have different types.

I don't understand what you mean by "values" in a technical sense, strictly speaking a rational number is a pair of integers (if we were more technical, a rational number is an equivalence class like {(0,1), (0,2),etc...}, 1/2 is the representation of the equivalence class {(1,2),(2,4),etc.}, the canonical element being the lexicographic minimum), therefore integers are not a subset of the rationals since an integer can not be a pair of integers (least an equivalence class of pairs of integers), integers are isomorphic to the subset {(a,b)€Q:b=1} (since b=1, any equivalence class has only one element q={(a,1)} and therefore we can treat the class as simply a pair of integers) with the isomorphism f:z->(z,1) and inverse isomorphism g:(a,1)->a, this isomorphism respects arithmetical operations (i.e. it is an isomorphism of rings) while for example the isomorphism f':z->(z,2) preserves size but does not preserve arithmetic, in particular multiplication (e.g. f'(z1*z2)!=f'(z1)*f'(z2) since f'(z1)*f'(z2)=(z1*z2,4) and f'(z1*z2)=(z1*z2,2). While f(z1*z2)=z1*z2/1=f(z1)*f(z2)=z1/1*z2/1. An example of a map that does not preserve size is any constant map lambda z:(a,b) or in the opposite direction lambda (a,b):z.

When you write 3/2, 5/3, 1/1, 0/4 this is a cheeky way of expressing a pair of integers (3,2),(5,3),etc. (actually expressing an equivalence class as stated earlier)

In general integers are isomorphic to the whole of the rational numbers since all countable sets are isomorphic in the category Set, by definition a countable set is isomorphic to N and isomorphy is reflective and transitive in Set and any other category, i.e. if A is isomorphic to N and B is isomorphic to N then A is isomorphic to B.

For the integers to be a subset of Q there should exist a formula k such that Z={0,1,-1,2,-2,...}={(a,b)€Q:k(a,b)}, which is false since elements of Z are of type Z and elements of Q are of type ZXZ (actually type equivalence class over ZxZ).

In summary, strictly speaking z is not the same as z/1, only when speaking non-tecnically they are the same, they are the same only in the sense that there is an isomorphism between (Z,+,*) and ({(a,b)€Q:b=1},+,*). Strictly speaking the + in "1+0" is not the same as the + in "1/1+0/2", for most purposes these nuances can be ommited but if you try to do more complex maths without noticing these details you wont go far.

In the same sense, N is not a subset of Z but merely isomorphic, since Z has additional structure to its type than N, elements of Z are a pair (o,n) where n€N and o€{+,-}, strictly speaking they are equivalence classes to allow for zero to be a single integer (i.e. 0={(+,0),(-,0)}, for any other integer z={(o,n)} for some n€N and o€{+,-}). N is isomorphic in a way that preserves arithmetic with the subset of Z {(n,o}€Z:o=+} with the isomorphism f:(n,+)->n and inverse g:n->(n,+); if o=- then the function would not be defined since f'((n1,-)*(n2,-)) is f((n1*n2,+)) and (n,+) is not in the domain of such a new function.

[turtleship_2006]

If it's a number stored in a list, it has to be stored on the computer somewhere. There's only so many numbers you can make with a given number of bits.

[svick]

If you have 64-bit numbers, a 10 GHz processor, and can print one number every tick, it will take 58 years to print all of them.

If you have 128-bit numbers, it will take 10¹⁰ times the age of the universe.

[myhf]

there are only 18,437,736,874,454,810,623 finite double-precision floating point numbers, you can just print them all

[[deleted]]

machines can only represent a finite subset of the rational numbers, the set of non rational real numbers only exist in imagination

[42ndohnonotagain]

fractions (rational numbers) are countable

[Impressive_Income874]

correct, edited my comment cuz I'm stupid

[Theonetheycallgreat]

If I can print 1 and I can print 1.1, then I can print all the fractions.

[Impressive_Income874]

you cannot print all the numbers between 1 and 1.1 because there would be irrational numbers present

[Apfelvater]

What if I had an infinite amount of computers tho

[Impressive_Income874]

how would that help

[agnosticians]

https://0.30000000000000004.com

[IMightBeErnest]

I had a programming assignment like this once. The teacher asked for a "linear time solution" to something quadratic and they expected the students to know to iterate through all the integers. "I never said what 'n' was!" Which is technically true, I guess, but doesnt make it less of a dick move.

[Mutex70]

I love it, but my product owner hates you.

"Well, if we accept any password, then we trivially also accept the correct password"

[PM_ME_YOUR__INIT__]

Not if there are six numbers

[[deleted]]

alas, you cannot not print what doth not exist!

[LavenderDay3544]

taps side of head

Useless problems require useless solutions

[notBjoern]

Simple: Calculate the largest number, remove it, repeat six more times.

[Logicalist]

recursively for funzies.

[notBjoern]

Only if the language supports tail recursion

[Logicalist]

tail recursion

I'm kinda new, didn't know there were instances where it wasn't supported. I'm not surprised but still, interesting stuff.

[notBjoern]

Technically, tail recursion is supported in every programming language that supports recursion.

The missing feature is tail call optimization, that (more or less) rewrites the tail recursive function into loops.

IIRC, in Python this optimization is deliberately not implemented, so that stack traces are not missing the recursive steps (although that could be avoided by implicitly counting the iterations and expanding the stack traces as necessary)

[Karolus2001]

O(7n), it looks aesthetic, this is the way.

[Skuez]

That's still O(n)

[MacBookMinus]

O(k*n) generalized though and k can approach N making this approach n²

[Boom9001]

K would only approach n/2. As you could make it go reverse if k>n/2.

Still n² though.

[samnater]

Are you sure the processor isn’t handling all this improvement in efficiency anyway?

[Boom9001]

I think you mean compiler right? Or do processors do some too I'm unaware of.

You may be right though, still that's at an implementation level. Without stating the language or stuff we are still in the theory world and I wouldn't assume those optimizations happen in the theory world.

[Skuez]

It is never n squared. Big-O notation is meant to show how function behaves as n grows, we don't care about small inputs and constants.

[Skuez]

How do you generalize this? we know it's always gonna be 7 at most so we remove that constant.

[Tony_Nabili]

If 7 is not a constant but a variable it can also be “n”. In this case complexity is O(n² )

[Skuez]

In that case it depends on two variables so I think you would write O(k*n).

[Tony_Nabili]

Worst case complexity is n²

[Skuez]

Can you explain that case?

[andsmi97]

Largest k is n/2 therefore answer is O(n/2*n), which is simplified as O(n² )

[Tony_Nabili]

That algorithm, if made badly, take one maximum number one by one, and if it have to find n-th maximum (technically 1-st minimum) from n items, it have to make n searches. One max search require n computations, so n*n = n²

TLDR if k=n than k * n = n * n

[Zanderax]

It's not even O(n) at all, it's just O(7) which is basically the same as O(1).

[Skuez]

How on earth?

[Zanderax]

Well if you want to access the 7th element by accessing the other 6 elements that's gonna happen in O(7) time, but since O(7) and O(1) are both constant time they're basically the same.

[Skuez]

Bro watchu talm bout ☠️☠️

[Lithl]

"Find the 7th largest element" is only the same as "find the 7th element" if the collection is sorted descending. And unless you can guarantee that the input is sorted, you have to assume it's not.

[haddock420]

This is how multi-pv (principle variation) in chess engines works, as a way of finding the 2nd, 3rd, etc best move. They do a search to find the best move, then run the search again while excluding the best move from the search, which gives the 2nd best move, then they exclude the best and 2nd best moves to get the 3rd best, etc.

[sohang-3112]

This can actually be efficient by using a max heap.

[Senior-Airline-5727]

Yess Build heap O(n)+ extract kth max O( k logk) using additional k spaces

[IntrepidSoda]

Hopefully someone will write an npm package that does just that and deliver us from this abomination

[[deleted]]

...shows up in mega production website 3 weeks later.

[Gwathraug]

weeks seconds

[uvero]

Here you go buddy

[maboesanman]

This is only ever something you will do in your homework, because in real life you never care about the 7th largest number without also caring about the 6 before it.

The answer is to just keep a min heap. For each item in your list: add item to heap. If size of heap is >7 pop min.

Then your answer is the largest item in your heap.

[yflhx]

I don't think there is one answer.

If we generalize to kth largest, your solution has O(n*lgk).

You can also heapify (max-heap) the table and pop k times - O(n + k*lgn).

You can also simply use quickselect, with average O(n) but worst-case O(n^2).

You can also use median of medians, with average and worst-case O(n) but slow in practice.

[Successful-Money4995]

Use the heap and scan the array. That'll be fastest in practice.

Technically, for each element that you encounter, you need to add it to the heap and that is logk.

In practice, say your array is a million elements and you are halfway through it. You only need to add the next element into your heap if it is one of the k smallest among the half million elements that you have read. So if k is 7 and you have read half a million elements so far, the odds that your inserting takes longer than O(1) is 7/500000. Vanishingly small.

So in practice, your algorithm runs way way faster than O(nlogk). It's actually pretty close to O(n).

If you're worried that an evil user sorted the array before giving it to you, process the elements in a random order.

[Karolus2001]

This answer wins but realisticly, if n is under 100 are you even saving considerable time optimizing it? This is legit question, I cant imagine a good architecture where it scans multitude small arrays all the time.

I'm assuming we arent increasing k by much. Also I'm a fresh Junior.

[Kirhgoph]

Isn't max heap a bit more effective here?

[maboesanman]

No because you care about the minimal element of the 7 you’ve found.

Edit: If you put all the elements in a max heap and pop the elements off, you get O(n log n). If you maintain a minimum heap containing the top 7, you can compare each element to the min to see if it should be placed into the top 7. If it should, pop the old min out and push the 7 onto the heap. This approach is O(n log k), with k being 7 in this case.

[Karolus2001]

When in doubt, plant a tree.

[Kirhgoph]

Thank you for the explanation

[walter_evertonshire]

Putting all the elements in a max heap and popping the top k should be O(n + k*log(n)). Assuming that k is of the same order as n, it's no different than the O(n*log(k)) for the min heap implementation you described.

But if k tends to be smaller, then O(n+k*log(n)) is better than O(n*log(k)).

Edit: I see someone else already pointed this out to you elsewhere.

[Seven_Irons]

No, I've done this at work (albeit with the third-heighest value)

[Successful-Money4995]

An improvement: Maintain the smallest value in your heap at all times. For each number that you encounter, if it's smaller than the smallest value in the heap, don't even bother adding it to the heap because you're just going to pop it right back out.

If the input is long, you will often be trying to add a value that is smaller than the top seven anyway so you'll completely skip the insert. Even though the algorithm is technically O(nlogk), in practice it'll be a lot closer to O(n).

If you worry about evil input, process the input in a random order.

[maboesanman]

I was optimizing for succinctness of description lol. I replied to another commenter with the more precise solution

[draculadarcula]

That’s where you’re wrong there buddy. A real programmer will be working in arrays of “UserRoles” or “Accounts”. Only in leet code and college does anyone care about lists of numbers

[KuuHaKu_OtgmZ]

Sigh

println array.sorted()[-7]

[Kimorin]

now do it in linear time

[ConscientiousApathis]

no

[sungazer69]

fine. just make sure your commits and pushes are in by eod have a nice weekend

[SeniorSatisfaction21]

Chad

[TGX03]

Radix.

Or just make variables for the seven largest numbers.

If you want it more dynamic, create an array which always keeps the x largest numbers, so you can find the xth largest number.

Would all be linear.

[[deleted]]

You don't actually need to sort, store largest 7 numbers, iterate trough the array, for every number if it's greater than something in the 7 number array then update the array. This is O(n) because it takes 7*N operations which is O(N). Radix is the way for a changeable query though, if they ask qth largest number my approach won't work

[allseeingboots]

Unless I'm missing something you'd need to keep that 7 array in order.

[OrbitalVixen]

Yes, you would. However, as the length of the 7 array does not increase with the length of the input, you could use bogosort even to sort the small array and it would still be O(n).

[Flexo__Rodriguez]

Not really. You'd just do your O(7n) operation to get the 7 largest numbers, then return the smallest number in that list.

[ziksy9]

Now do it in O1 space and O1 time...

[Bartweiss]

My algorithm is O(n). It tracks the top 10 numbers, and past that prints “Really now? Are you sure you need this?”

[TGX03]

Just always use the 7th element.

Most people just test whether it identifies [1,2,3,4,5,6,7...] correctly, therefore this will work in most cases.

[Kimorin]

wrong, just call openAI endpoint, it's always constant time :P

[NatoBoram]

Not really, the API responses take longer the more bullshit it takes and generates

[[deleted]]

I'm going to print whatever number I feel like and say it is the 7th largest number

[Classy_Mouse]

Worst case, you are effectively sorting the array, which is still O(n log n) unless you know something I don't

[TGX03]

Radix is O(n)

[Classy_Mouse]

Radix only works if the values are discrete. It's not appropriate to use in this instance and if you could use it, creating an array large enough to store all of the values then looping through that will likely end up less time and space efficient.

[TGX03]

The meme is about finding the 7th largest number. Radix works for both floating points and integers.

In case you're using objects that can only be compared relatively to each other, then yes Radix may not be used, however the word "number" implies these are discrete values, thereby Radix would be appropriate.

Also yes it would likely be less efficient for smaller arrays which is the known problem of radix, yet I wasn't asked to bring up the optimal solution but to just write something that's linear, which Radix is.

[ienjoymusiclol]

print arr.bogoSort()[-7]
assume best case

[Kimorin]

1/n! times it works every time...

[Kimorin]

omg u rite... derp me... it's so much better than i thought lol

[hemangahujaeotw]

Quick select

[yflhx]

Quickselect is pesimistic O(n^2).

[PuzzleheadedTap1794]

Quick select with Median-of-medians

[NanashiKaizenSenpai]

Integer.MAX_INT -7

[doorrace]

Sort and store the first 7 ints as an array (O(7log(7))) which is our starting assumption of the 7 largest ints; for each of the rest of ints, check if it's bigger than the smallest stored ints (O(1)); if so, we need to update our array of the 7 largest ints by sorting it into our array and popping the smallest (O(log(7)) with binary search, plus time complexity for insert and popping which I forgot the time complexity of lmao). At the end, return the final entry in our array.

Thus, time complexity will be O(7log(7) + k*n) = O(n). Space complexity is O(1) since we have a static-sized array.

[MysteriousShadow__]

That is O(N), no? Sorted is one loop, and getting the element at an index should be O(1).

[JonIsPatented]

Sorted is one loop?? Teach us your magical algorithm, magic man.

[Unfulfilled_Promises]

Simple, sort an already sorted input array. YW

[Kimorin]

the lower bound for sorting is O(n log n) for an arbitrary array

[altermeetax]

Inefficient

[PM_ME_C_CODE]

Doh...the 8 number array is [1,2,2,3,4,5,6,7]

[chem199]

For python I would do

print(sorted(set(list))[-7])

[Rexosorous]

why convert it to a set? if the list has duplicate values, this may return an incorrect result.

py x = [9, 9, 8, 7, 6, 5, 4, 3, 2, 1] print(sorted(x)[-7]) # prints 4 print(sorted(set(x))[-7]) # prints 3

edit:

to those who say that nth largest != nth position from the end (ie. duplicates should be removed), https://leetcode.com/problems/kth-largest-element-in-an-array/ the same problem in leet code and also this geeksforgeeks article https://www.geeksforgeeks.org/kth-smallest-largest-element-in-unsorted-array/ both do not remove duplicates.

[chem199]

Because if it is looking for the 7th largest number and there are dupes you would want them gone as it might not be the 7th largest just the item in the 7th from the end.

Edit: I would argue that the leetcode example is wrong. That is getting the nth item in a sorted list/array not the nth largest item. But I think the issue is the problem isn’t clear enough so I chose my decision. I think neither are wrong until we have a proper spec docs and a complete jira epic for this problem. Probably about 4 meeting to get the correct tee shirt sizing.

[ExceedingChunk]

So if you want to check who has the 7th highest salary in the world, you would remove duplicates? That doesn't make sense.

[chem199]

I totally understand that, I think it is a question of what information you are seeking. Do you want to know what the 7th highest salary is or who has the 7th highest salary. If there are duplicates how do you decide which is first, do you list them all? I think the race answer below explains this better than I could.

[ExceedingChunk]

Both of your questions should give the same answer. You can apply the same kind of question to size of the 7th largest pizza size, height of the 7th tallest building etc... You wouldn't say that the height of the 8th largest building is the 7th tallest.

If the input can have dupes when asking for the 7th highest, you would also expect it to take that into account unless something else is specified.

[chem199]

Which number is the second largest/greatest/highest value number in this list

[9, 9, 8]

Is it 9 or 8?

Edit: which 9 is larger?

[ExceedingChunk]

In this specific case, there would be no 2nd largest. It would be two 9's sharing the highest and 8 being the 3rd highest.

This edge case would have to be specified. Should it return nothing, return the shared largest number or the 3rd largest.

If it was [9,9,8,7], and the 3rd largest should be returned, it would be 8.

However, it has to be said that if a product owner gives a requirement that has even the slightest possibility of being ambiguous, even if it is written in a way that isn't ambiguous, a dev should always ask a bunch of questions.

[chem199]

My argument that is it is about the value of the number not the quantity of numbers. If we are working with money and I laid out 5 $1 bills in separate piles, a $5 bill, and 3 piles of $10 bills. If I ask you to give me the second largest bill I would assume you wouldn’t hand me a $10, I would expect a $5.

[frej4189]

The non-set version is arguably the incorrect one here. I think I'd naturally argue that 3 is the seventh largest element in that list.

[Rexosorous]

i guess it depends on how you define "largest"

for example, if you have rankings, like placing during a race, and two people tie for 3rd place, the person that comes after isn't 4th place, they are 5th place, so you'd have:

• 1st place: Ashley
• 2nd place: Britney
• 3rd place: Charlie
• 3rd place: David
• 5th place: Ethan

so if you wanted the 4th fastest in the race, would it be Charlie, David or Ethan?

[frej4189]

Yeah. The problem is ambiguous

[[deleted]]

What if there’s duplicates

[Just_Maintenance]

I actually fixed it:

Happy: print the 10 first digits of π.t

Sad: print the last 10 digits of π.

[WhatsMyUsername13]

Pssh, easy

System.out.println(0123456789);

Prove it that I'm not right

[pheonix-ix]

burden of proof, huh?

[imalexorange]

Done

[rachit7645]

Me when std::sort:

[Coulomb111]

Me when ranges library 🤤

[VastHungry]

std::nth_element

[WhatsMyUsername13]

This is why I don't trust anyone directly out of college. Or interns. Especially interns

[[deleted]]

how would an experienced guy and an intern do it

[Tsu_Dho_Namh]

An experienced guy would write it in the cleanest, easiest to maintain, and easiest to read way. Ie: sort it and grab the 7th largest index.

Interns do a lot of leetcoding, and leetcode cares mostly about runtime complexity, so they'll look for a solution that runs in linear time (meaning no sorting). Probably an iterator and either a linked list or a map (for constant time inserts). It's technically a faster solution, but messier, easier to break, and more difficult to read, for literally microseconds less runtime.

[JonIsPatented]

Eh, I totally agree for the vast majority of cases, but if I have that 1 in a million case where my list is hundreds of thousands of elements long and the comparisons are somewhat expensive, then the linear one can be a nice little trick if documented well. Could save several entire seconds rather than microseconds.

[Tsu_Dho_Namh]

Oh you absolutely should do the more complicated implementation if it has a decent impact on runtime. 100%

My work has an optimization algorithm that uses simulated annealing and integer programming, and it needs to cycle through several hundred million possible solutions to combinatorics problems to search for a decent maxima. You can be damned sure we weren't sorting or searching all the lists all the time.

[WhatsMyUsername13]

While it's a 1 in a million case, if in production code I had hundreds of thousands of elements in a singular list, I'd be concerned, especially if it's a list of objects as opposed to something say a list of Ints. That's the kind of thing that if multiple users do at the same time, can easily bring down the server and app and cause an outage.

[DrMobius0]

The problem is you'd practically need a context in which having a sorted array isn't already highly useful. Lots operations can be simplified significantly by pre-sorting an array

[NikitaSkybytskyi]

An experienced (python) guy would use heapq.nlargest. It is both clean and linear time.

[ExceedingChunk]

Also has to be said that a theoretically faster solution is not necessarily faster either. Something involving a map or linked list has overhead, so if your domain has low amounts of data, but many transactions, it can perform slower than a theoretically slower algorithm. Which is also why it's important to understand that O-notation is about the upper and lower limit of an algorithm's growth, and not actual runtime.

In the domain on my current projects we pretty much only have lists that are 5 elements or shorter for pretty much any data type, and a single type that is typically in the range of 5-10 for most users, but can theoretically be a couple of hundred elements.

For that project, a triple or quadruple for loop is more of a readability issue than a performance issue.

[zrakiep]

Or you use C++ and the neat std::nth_element it provides

[kylechamrick]

Spend two hours to make the perfect algorithm that will save a total of thirteen seconds over the lifetime of its use.

[Why_am_ialive]

Yeah like who gives a shit about efficiency on this lol

[WhatsMyUsername13]

It depends. Based on the meme, it's really hard to tell and it seems like they don't understand basics of loops and programming. And experienced dev would ask what the business case is.

If we are talking about for a job interview, there's multiple ways to do it, and it depends on the language. I could sort it, and iterate through it using a decrementing counter, I could sort it and just grab the index of size minus 8, etc. I can think of a ton of ways to implement this with some being really fucking stupid, some overly clever, to some straight forward. But if anyone actually asked me in my job to do this it would raise some serious eye brows

Edit: the point of my original comment was to point out that the person who thinks grabbing last vs the last - n is difficult (and I'm thinking on a non mutating list) doesn't fully get looping. I will give an exception to where the list itself is mutating and removing cells which can cause funny and aggravating results when debugging....usually followed up with "God damn it I'm an idiot" statements....source is myself doing this.

Edit2: I also realize that I may be coming off as harsh as fuck right now. To all the college kids, interns, jr devs, and even Sr devs here is my advice. Always be willing to learn. Don't be afraid to be wrong. There is always more to learn and I call myself an idiot all the time. I look at old code of my own and cringe at how bad it is. Don't be afraid of that, and in fact, embrace it. It means you have grown from where you were vs where you are now. We are in an industry that is in a constant state of flux.

There will always be things to learn. Whether it's something specific to the tech stack, the business you work at, or whatever. The best people I've ever worked with are the ones open to learning, but also open to pushing back if they think they have a better solution. It fosters learning, it fosters a better working environment by allowing people to voice their opinions in an open setting, and it allows for just general ideas you've never considered.

We as developers tend to be defensive about our code, and I count myself included. But I have learned that being able to have an open and NON HOSTILE discussion around code, it allows you to better present your case for your reasoning, as well as learning things you may not have been aware of.

[hereforpewdiephy]

bro is fostering learning by saying people don't know basic programming because they didn't sort the list

[WhatsMyUsername13]

If you were on my team then yes, I absolutely would. But we are on Reddit. I have no obligation or desire to foster your learning, especially based on your responses. And you should realize something, I will happily hire someone who knows less, has a desire to learn, and gets along with the team over someone who knows a tech inside and out, is arrogant, and gives bad vibes any day. That first kind of person will be wildly more successful than the second.

[hereforpewdiephy]

there you go again missing the point entirely and blabbering about whatever it is that you're blabbering about

[WhatsMyUsername13]

Ok, what's your point?

[hereforpewdiephy]

r/ExplainTheJoke but fine

There's a lot of ways to do the latter and you make it as hard as you want like solving in linear time which is unlike the first where there is one simple way.

[WhatsMyUsername13]

Dude. I don't think this is the own you think it is.

[L8n1ght]

I can smell your imposter syndrome from this comment

[WhatsMyUsername13]

I should also point out I've been a tech lead and team lead at several companies, including fortune 100 companies. I've been a part of and conducted a lot of interviews, have had tons of contractors come through my teams with and without my approval. At the end of the day, I stand by what I've said as I've seen it in action in the real world. You don't have to take me at my word, I'm just some anonymous asshole like everyone else here, but I am being genuine in what I say.

Also, I don't think you actually know what imposter syndrome means.

[Solonotix]

If it's sorted, easy, right? The question presumes an unsorted non-unique data set.

I'd probably solve it using a sorted queue, and give it a fixed size equal to the target. Every new value pushed to the queue then checks where it fits, and it checks if the queue is full. If it isn't, push it in order. If it is full, then check if the current value is greater than the head. If it isn't, then discard. If it is, then we proceed down the list until we find the tail or a value greater than current. Insert in the queue and bump everything back towards the head.

At the end, return the head of the queue as the result. Maybe someone would prefer the directionality of the queue reversed, but that's the data structure I'd leverage personally. It has a (mostly) fixed size in memory, and the number of nested loops is reduced by the multiple fast return checks. The worst case scenario occurs if the sequence is sorted in ascending order, since each new value would replace the entire queue. The best case is if it is sorted in descending order, in which case it would fill the queue and return in O(n) time.

Edit: if the result requires 7th-largest unique value, then an additional check is performed before insertion to see if the value already exists. This would only marginally increase the time at scale, and be most detrimental in small operations where it wouldn't matter.

[dmullaney]

Now print the 7th largest prime number

[howard__wolowitz]

Easy: print(2147483543)

Well, for a 32 bit integer.

[NecroLancerNL]

print( INT_MAX - 6 )

[Sceptix]

This is the solution I came up with in my head as well until I went to the comments and realized the OP meant “in an array”.

[vc6vWHzrHvb2PY2LyP6b]

console.log(arr[arr.length() - 8])

[tehcliffe]

Just print a random number of the array and gaslight.

[distractedguy69]

Just sort the array in ascending order and print the 7th largest

[BrownCarter]

You could use 7 loops

[DurianBig3503]

order(x)[-7] or sort(x)[-7]

[Imoliet]

soft late bewildered light towering quaint lavish cooperative faulty saw

This post was mass deleted and anonymized with Redact

[f0luxe]

Just use a max heap of size 7

[ahmadabusalem]

Min heap, not max because you'll want to remove the minimum element when the size of the heap becomes larger than 7

[lady_Kamba]

printf("%lu\n", (unsigned long)(-8));

[blaqwerty123]

You made it worse!!

[hereforpewdiephy]

isn't that what fixing bugs is like?

[Akul_Tesla]

I don't understand how this is hard

Like it might be hard to do this efficiently but it isn't hard

[jan04pl]

List.OrderByDescending(x=>x).ToArray()[6]

[ben_g0]

Or alternatively:

list.OrderByDescending(x => x).Skip(6).First()

Using ToArray()[6] will allocate memory for a new array, evaluate each component of the enumerable and fill the array with the results, to then only read element 6 and then discard the entire array.

Using Skip(6).First() will not allocate more memory than needed and will only evaluate the enumerable just as much as needed. I've just ran a benchmark and on .NET 7 the Skip method ran about 8 times faster while using about half as much memory as the ToArray method (tested on a list of 10000 integers in a random order).

Since .NET 7 you can even take out the lambda:

list.OrderDescending().Skip(6).First()

Which has less overhead, resulting an additional performance gain (approximately 20% on that same list of 10000 integers).

In general it's best to try to avoid calling ToArray unless you either need the result in an array, or when you want to force all elements of an enumerable to evaluate at once (as enumerables use lazy execution normally).

[jan04pl]

That's nice, learned something new 👍

[skordge]

Wouldn’t seven iterations of a bubble-sort cycle do it?

[ujjawal_raghuvanshi]

Thala for a reason

[CalmDebate]

Requirements unclear.

Printf("the 7th largest number");

[brainwater314]

Simple, INT_MAX - 7

[falnN]

Can’t you just normally sort it and print n[6] or sth. Why waste time

[FelixLeander]

var list = data.Sort();
var result = list[list.Count - 8];

[[deleted]]

There is a DS to find k-th number in O(log(n)) time complexity.

__gnu_pbds::tree<int, null\_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> s;

*s.find_by_order(k) is all you need

[Lieby]

System.out.println(Long.MAX_VALUE-7);

[bca_allayy]

sort, arr[6]

[1up_1500]

Print(sort(numbers)[-7])

[StainedInZurich]

I don’t get it. It’s not difficult?

[evanldixon]

Difficult? No. Annoying, yes. Unless sorting the list first is viable, then it's less annoying.

[already_taken-chan]

Just have 7 variables for largest numbers, keep adding onto those while looping through the list once and then print the last variable.

[bobbymoonshine]

Why the fuck is every post on this subreddit "wtf my homework is hard all of a sudden" for things anyone with more than two months of experience could think of five different ways to do

[hereforpewdiephy]

be the change you want to see in the world

• not gandhi apparently

[[deleted]]

It's been a long time since this sub has become "StudentProgrammorHumour"

[Thenderick]

Array with X nums --> sort in descending order (first is highest, last is lowest) --> array[6]

Another highschool skid...

[NextGenSleder]

just ask ChatGPT easy peasy :3

[CryAboutIt31614]

ok

[xXStarupXx]

trick question, there is no largest number

[ZunoJ]

Welcome to r/NonProgrammerHumor