There's a famous question posted on this site which asks about finding the $k$th largest element. Many answers are written there which optimized it and found algorithms with expectation of $O(n)$.

The thing I don't understand is.... Those algorithms wouldn't work if $k$ is dependent on $N$. Therefore $k$ must be a constant.

But if it's a constant why isn't it easiest to simply loop like bubble sort for $k$ times? So you pushed the $k$ largest elements to the end of the array. No? Complexity would be $O(nk)$ but $k$ is a const so it would be $O(n)$.

Why do we need crazy algorithms like median of medians and using quicksort if this alone works?

  • $\begingroup$ Perfectly! Finding each Max take O(n). Then all you have to do is repeat the process k times. Since k isn't dependent on N wouldn't it be considered O of n? This looks too simple so why do people use crazy algorithms like quicksort and stuff? $\endgroup$ – bilanush Nov 6 '18 at 1:42
  • $\begingroup$ The key point is exactly that $k$ may dependent on $N$. For example, how can you find the median? Here $k$ is $N/2$ or $(N+1)/2$. Those "crazy" algorithms work for that case. $\endgroup$ – John L. Nov 6 '18 at 1:53
  • $\begingroup$ Mmm I thought about that. But i would think if k isn't constant but n dependant. Then.... Even using those 'crazy algorithms' wouldn't help much either. As they too would run in n squared or nlgn times. For example if you use a modified version of quicksort as suggested here en.wikipedia.org/wiki/… then... I don't see any reason it wouldn't take n squared like quicksort in worst case. So then again you haven't got any advantage compared to my simple algorithm $\endgroup$ – bilanush Nov 6 '18 at 2:23
  • $\begingroup$ I am afraid you do not understand those algorithms. For one example, please check my answer to Median of medians: bound on pivot position. $\endgroup$ – John L. Nov 6 '18 at 2:30
  • $\begingroup$ Ok I really do not understand them so well. I would just rely on you. Just tell me if this is correct. What you basically say is... That even if k isn't constant but grows as n grows these algorithms wouldn't take more than O(n), right? Even though quicksort itself does take a higher order $\endgroup$ – bilanush Nov 6 '18 at 2:33

The $k$th largest element can be found in time $O(n)$ for all $k$ using a deterministic algorithm. See Wikipedia or many textbooks, such as Cormen et al., Introduction to Algorithms.

Given the $k$th largest element, you can find the $k$ largest elements in $O(n)$ using a simple scan. If all elements are distinct, you just output all elements which are at least as large as the $k$th largest. Without this assumption, you output all elements which are strictly larger than that element, and then enough copies of the $k$th largest one.

  • $\begingroup$ Sorry I might be mixing things up. Maybe what I am actually trying to achieve is finding all k largest elements and not just the Kth one. Sorry for not being clear. I wrote it wrongly. Wouldn't it still be O of n if I go about finding largest elements . Each search for Max takes O of n and I need to iterate it k times. So I got K*n. So why do people offer much more complicated algorithms for this problem? $\endgroup$ – bilanush Nov 6 '18 at 1:40
  • $\begingroup$ Even for just finding the Kth one. I am sorry but I don't understand why does your Wikipedia page go so far with crazy algorithms? I have a much simpler one. As I noted in previous comment and in the question $\endgroup$ – bilanush Nov 6 '18 at 1:45
  • $\begingroup$ Once you've found the $k$th largest element, you can find all others in $O(n)$ by going over all elements and comparing them to the $k$th largest element. This assumes all elements are distinct; otherwise you'll have to work a tiny bit harder. $\endgroup$ – Yuval Filmus Nov 6 '18 at 1:49
  • $\begingroup$ Ok . But my algorithm is way simpler than the ones you presented... in Wikipedia. Why not use my simple algorithm? Just find k times the max $\endgroup$ – bilanush Nov 6 '18 at 2:18
  • $\begingroup$ Your algorithm takes quadratic time to find the median. Quickselect only takes linear time. $\endgroup$ – Yuval Filmus Nov 6 '18 at 2:20

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