quicksort

So choosing a pivot at random has O(n 2 ) running at worst case but when the pivot is chosen as the average of min value and max value of the list you get a worst case O(n log n). Of course there are the added 2*O(n) on each recursion due to finding the min and max values as opposed to the constant O(1) of the random generator has. When implementing this as the pivot you get the list sorted at the leaves of the recursion tree instead in the standard algorithm elements get sorted from the root to leaves. When implementing instead of the pivot being a value on the list it is just a number

In Worst case Quicksort recursion Depth requires Stack space of O(n). Why it doesn't cause a stack overflow for large set in the worst case? (reversed sequence) If you recurse on both sides of the pivot then it does cause stack overflow for sufficiently large data in the worst case. That's why nobody uses a naive QuickSort in production code. There is a simple change you can make to the algorithm to prevent Omega(n) worst-case stack use. After each partition, recursively Quicksort the "small half" and then iteratively loop around to do the "big half". This requires O(log n) additional stack

I'm reviewing algorithm stuff and stuck in a simple quick sort algorithm implementation in java import java.util.Arrays; import java.util.Random; public class QuickSort { int arr[]; boolean randomPick; Random rand = null; public QuickSort(int[] inputArray) { arr = inputArray; randomPick = false; } public QuickSort(int[] inputArray, boolean random) { arr = inputArray; if (random) { randomPick = true; rand = new Random(System.currentTimeMillis()); } else { randomPick = false; } } public int[] sort() { int start = 0; int end = arr.length - 1; try { _sort(start, end); } catch (StackOverflowError e

This question already has answers here : is it possible to do quicksort of a list with only one passing? (3 answers) Why is the minimalist, example Haskell quicksort not a “true” quicksort? (11 answers) Closed 5 years ago . Here's an example from Learn you a Haskell: quicksort :: (Ord a) => [a] -> [a] quicksort [] = [] quicksort (x:xs) = let smallerSorted = quicksort [a | a <- xs, a <= x] biggerSorted = quicksort [a | a <- xs, a > x] in smallerSorted ++ [x] ++ biggerSorted It seems that the list is iterated twice for each recursion, once for each list comprehension. Is there some magic in the