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randomized select not give steady solution

EDIT
try run in the main:
int[] arr = {646 ,94 ,366 ,754 ,948 ,678 ,121 ,320 ,528 ,36};
for(int i=0;i<10;i++){
System.out.println(randomizedSelect(arr,0,arr.length-1,5));
printArr(arr);
}
and see that i got diffrent outpot in each loop..
Got a little problem that I would like some help with, if anyone knows how.
I need to find the kth smallest value in an array by randomized partition.
I've got two problems:
I get array out of bounds with -1 and can't find a way to fix it.
Most of the time it works but sometimes it gives me wrong k place.
For example for array with length of 10, it tells me that 20 is in the 5th place but actually it should be in the 2nd place and it prints the array where not all the values on the left are smaller than 20 and not smaller than the 5th place.
Here is an example array:
{646 ,94 ,366 ,754 ,948 ,678 ,121 ,320 ,528 ,36}
The array input is done by a random number generator.
This is my code:
import java.util.Random;
import java.util.Scanner;
public class Main {
static Scanner scan = new Scanner(System.in);
static Random rand = new Random();
public static void main(String[] args) {
int nSize = askSizeN();
int kSize = askSizeK(nSize);
int[] arr = new int[nSize];
chose(arr);
int[] arrCopy = new int[nSize];
for (int i = 0; i < arrCopy.length; i++) {
arrCopy[i] = arr[i];
}
printArr(arrCopy);
System.out.println(randomizedSelect(arrCopy, 0, arr.length - 1, kSize));
printArr(arrCopy);
}
private static int partition(int[] arr, int p, int r) {
int x = arr[r];
int i = p - 1;
for (int j = p; j < r; j++) {
if (arr[j] <= x) {
i++;
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}
int temp = arr[i + 1];
arr[i + 1] = arr[r];
arr[r] = temp;
return i + 1;
}
private static int randomizedPartition(int[] arr, int p, int r) {
int i = rand.nextInt(r - p);
int temp = arr[r];
arr[r] = arr[i];
arr[i] = temp;
return partition(arr, p, r);
}
private static int randomizedSelect(int[] arr, int p, int r, int i) {
if (p == r) {
return arr[p];
}
int q = randomizedPartition(arr, p, r);
int k = q - p + 1;
if (i == k) {
return arr[q];
}
else if (i < k) {
return randomizedSelect(arr, p, q - 1, i);
}
else {
return randomizedSelect(arr, q + 1, r, i - k);
}
}
private static int askSizeN() {
System.out.println("Please chose the size of the heap: \n" + "(the size of n)");
return scan.nextInt();
}
private static int askSizeK(int nSize) {
System.out.println(
"Please chose how much small values you want to see: \n" + "(the size of k)");
int kSize = scan.nextInt();
if (kSize > nSize) {
System.out.println("cant print more number then the size of the Heap..");
System.out.println("Please enter a number less then " + (nSize + 1));
askSizeK(nSize);
}
return kSize;
}
private static int[] chose(int[] a) {
System.out.println("Chose the option you want: \n" + "\t1. enter your own values."
+ "\n\t2. let me generate random values");
int chose = scan.nextInt();
if (chose == 1) {
for (int i = 0; i < a.length; i++) {
System.out.println("Enter value number " + (i + 1));
a[i] = scan.nextInt();
}
}
else if (chose == 2) {
System.out.println("Generate random numbers.");
for (int i = 0; i < a.length; i++) {
a[i] = rand.nextInt(1000);
}
}
else {
chose(a);
}
return a;
}
private static void printArr(int[] a){
for(int i=0;i<a.length;i++){
System.out.print(a[i] + " ");
}
System.out.println();
}
}

I've solved the problem.
Method randmizedPartition() was generating wrong random pivot for partition.
I solved it by changing the random line to:
int i = rand.nextInt((r - p) + 1) + p;

Java QuickSort Best Case Array Generation

I've been banging my head on the table on this one.
I need to create an n sized array that is optimized for QuickSort Partition. It will be used to demonstrate the growth of QuickSort's best case. I know that for best case, QuickSort must select a pivot that divides the array in half for every recursive call.
I cannot think of a way to create an n-sized optimized array to test. Any help would be greatly appreciated.
Here is the algorithm in Java.
public class QuickSort {
private int length;
private void quickSort(int[] a, int p, int r) {
if (p < r) {
int q = partition(a, p, r);
quickSort(a, p, q - 1);
quickSort(a, q + 1, r);
}
}
private int partition(int[] a, int p, int r) {
int x = a[r];
int i = p - 1;
for (int j = p; j < r; j++) {
if (a[j] <= x) {
i++;
exchange(a, i, j);
}
}
exchange(a, i + 1, r);
return i + 1;
}
public void exchange(int[] a, int i, int j) {
int tmp = a[i];
a[i] = a[j];
a[j] = tmp;
}
QuickSort(int[] a) {
if (a == null || a.length == 0) {
return;
}
length = a.length;
quickSort(a, 0, length - 1);
}
}

I know this is an old question, but I had the same question and finally developed a solution. I'm not a Java programmer, so don't blame me for Java code issues, please. I assumed that the quicksort algorithm always takes the first item as a pivot when partitioning.
public class QuickSortBestCase
{
public static void generate(int[] arr, int begin, int end)
{
int count = end - begin;
if(count < 3)
return;
//Find a middle element index
//This will be the pivot element for the part of the array [begin; end)
int middle = begin + (count - 1) / 2;
//Make the left part best-case first: [begin; middle)
generate(arr, begin, middle);
//Swap the pivot and the start element
swap(arr, begin, middle);
//Make the right part best-case, too: (middle; end)
generate(arr, ++middle, end);
}
private static void swap(int[] arr, int i, int j)
{
int t = arr[i];
arr[i] = arr[j];
arr[j] = t;
}
private static void fillArray(int[] arr)
{
for(int i = 0; i != arr.length; ++i)
arr[i] = i + 1;
}
private static void printArray(int[] arr)
{
for(int item : arr)
System.out.print(item + " ");
}
public static void main(String[] args)
{
if(args.length == 0)
return;
int intCount = Integer.parseInt(args[0]);
int[] arr = new int[intCount];
//We basically do what quicksort does in reverse
//1. Fill the array with sorted values from 1 to arr.length
fillArray(arr);
//2. Recursively generate the best-case array for quicksort
generate(arr, 0, arr.length);
printArray(arr);
}
}
This program produces the same output for the array of 15 items, as described here: An example of Best Case Scenario for Quick Sort. And in case someone needs a solution in C++:
template<typename RandomIterator,
typename Compare = std::less<typename RandomIterator::value_type>>
void generate_quicksort_best_case_sorted(RandomIterator begin, RandomIterator end)
{
auto count = std::distance(begin, end);
if (count < 3)
return;
auto middle_index = (count - 1) / 2;
auto middle = begin + middle_index;
//Make the left part best-case first
generate_quicksort_best_case_sorted(begin, middle);
//Swap the pivot and the start element
std::iter_swap(begin, middle);
//Make the right part best-case, too
generate_quicksort_best_case_sorted(++middle, end);
}
template<typename RandomIterator,
typename Compare = std::less<typename RandomIterator::value_type>>
void generate_quicksort_best_case(RandomIterator begin, RandomIterator end)
{
{
auto current = begin;
RandomIterator::value_type value = 1;
while (current != end)
*current++ = value++;
}
generate_quicksort_best_case_sorted(begin, end);
}

How to avoid StackOverflowError caused by the number of recursion calls in Java?

I want to solve this problem: https://www.hackerrank.com/challenges/find-median ,i.e. find the median element in unsorted array. To do this I perform quickselect algorithm.
My program works correctly on my computer. However when I submitted in the system it gave me StackOverflowError. I think this is because of the depth of the recursion calls. I guess I make too many recursion calls than Java allows (the error is caused by a test case with 10 001 numbers). Can someone suggest me how to avoid this?
This is my code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class FindMedian {
static int res;
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(
new InputStreamReader(System.in));
String line1 = br.readLine();
int N = Integer.parseInt(line1);
String[] line2 = br.readLine().split(" ");
int[] arr = new int[N];
for (int i = 0; i < N; i++) {
arr[i] = Integer.parseInt(line2[i]);
}
selectKth(arr, 0, N - 1);
System.out.println(res);
}
public static void selectKth(int[] arr, int start, int end) {
// it is written to select K-th element but actually
// it selects the median element
if (start >= end) {
res = arr[start];
return;
}
int pivot = arr[start];
int n = arr.length - 1;
int i = start + 1;
int j = end;
while (i <= j) {
while (arr[i] <= pivot) {
i++;
}
while (pivot < arr[j]) {
j--;
}
if (i < j) {
int tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
i++;
j--;
}
}
int tmp = arr[j];
// j is the index of the last element which is bigger or equal to pivot
arr[j] = pivot;
arr[start] = tmp;
if (n / 2 <= j) {
selectKth(arr, start, j);
} else {
selectKth(arr, i, end);
}
}
}

A basic concept of IT science is to replace recursion with iteration. By using iteration you'll never run into "recursion too deep" errors. Every problem can be solved either by recursion or by iteration.
See my link for details.
http://www.refactoring.com/catalog/replaceRecursionWithIteration.html

Since you only call selectKth at the very end of itself (it's practically tail recursive already), it's trivial to unroll your recursion into an iteration.
func foo(param) {
if (param == bar) //bailout condition
return;
x = doStuff(param); //body of recursive method
foo(x); //recursive call
}
Can be rewritten as
func foo(param) {
x = param;
while ( x != bar ) {
x = doStuff( x );
}
}
In your case the only tricky bit is the last if:
if(n/2 <= j){
selectKth(arr, start, j);
}
else{
selectKth(arr, i, end);
}
So your method will look like this:
public static void selectKth(int[] arr, int start, int end) { //it is written seclect K-th element but actually it selects the median element
while( start < end ) {
//method body as is up to the last `if`
if(n/2 <= j) { //instead of recursion we just adjust start and end, then loop
end = j;
}
else{
start = i;
}
}
res = arr[start];
}

Why do I get a stack overflow error?

Here is the code .. I have to sort an already sorted array and to calculate it's execution time ...for quicksort it is n^2 cause it is the worst case. but for large input data let's say 7500 it gives me an overflow error :S what can i do in order to calculate the running time?
public class Provo {
public static void swap(int[] a, int i, int j) {
int temp = a[i];
a[i] = a[j];
a[j] = temp;
}
public static int HoarePartition(int m, int d, int[] a) {
int pivot = a[m];
int i = m + 1;
int j = d;
while (i < j) {
while (a[i] < pivot) {
i = i + 1;
}
while (a[j] > pivot) {
j = j - 1;
}
if (i < j)
swap(a, i, j);
}
swap(a, m, j);
return j;
}
public static void quicksort(int m, int d, int[] a) {
if (m < d) {
int s = HoarePartition(m, d, a);
quicksort(m, s - 1, a);
quicksort(s + 1, d, a);
}
}
}
and here is the main class
import javax.swing.*;
public class ascending {
public static void main(String[] args){
String input=JOptionPane.showInputDialog("Shkruani nr e te dhenave");
int size=new Integer(input).intValue();
int[] r= new int[size];
int[] p = new int[size];
int majtas=0;
int djathtas=size;
for(int i=majtas;i<djathtas;i++)
{r[i]=i;}
for(int i=majtas;i<djathtas;i++)
{p[i]=r[i];}
long average;
int n=100;
long result=0;
for(int j=1;j<=n;j++)
{
long startTime = System.nanoTime();
Provo.quicksort(majtas,djathtas-1,p);
long endTime = System.nanoTime();
result = result+(endTime-startTime);
long a = endTime-startTime;
System.out.println(j+": " +a);
for(int i=majtas;i<djathtas;i++)
{p[i]=r[i];}
}
average=result/n;
System.out.println("Koha e ekzekutimit te insertion sort eshte " + average + " nanosekonda ");
}
}

Well if you are stuck, you can maybe look for iterative quicksort on the web to get some help.
I have found this article. It is dedicated to C# but translating it into Java shouldn't be a big issue.

Change int djathtas = size - 1; to int djathtas = size; and change quicksort(majtas, djathtas, p); to quicksort(majtas, djathtas - 1, p);
Otherwise it's not allocating 10 digits, only 9.
It seems like you're just using too large of a number. Your program can be functioning correctly and generate a stack overflow error. You could try implementing another version of quicksort that does not rely on the stack or something.
Other than that, I'm not sure why you need such a large input anyway.

Quicksort: Almost sorted, but not quite. What's wrong?

Here's the code. The output is a very nearly correctly sorted array, but there are several elements out of order. Anyone able to spot the error?
I'm pretty sure the swap and quicksort methods are correct, but I'm posting all the methods here just in case.
package quicksort;
import java.util.Random;
import java.util.Arrays;
public class QuickSort {
/**
* #param args the command line arguments
*/
private static int[] u;
public static void main(String[] args) {
u = makeArray(100);
System.out.println(Arrays.toString(u));
quicksort(0, u.length - 1);
System.out.println(Arrays.toString(u));
}
public static int[] makeArray(int n) {
int[] a = new int[n];
int j;
Random r = new Random();
for (int i = 0; i < n; i++) {
j = (r.nextInt(100) + 1);
a[i] = j;
}
return a;
}
public static int partition(int left, int right, int pivot) {
int p = pivot; // pivot
int lPt = left - 1;
int rPt = right + 1;
while (true) {
while ((lPt < right) && (u[++lPt] < p));
while ((rPt > left) && (u[--rPt] > p));
if (lPt >= rPt) {
break;
} else {
swap(lPt, rPt);
System.out.println("Swapping " + lPt + " " + rPt);
}
}
return lPt;
}
public static void swap (int a, int b) {
int temp = u[a];
u[a] = u[b];
u[b] = temp;
}
public static void quicksort(int l, int r) {
if (r - l <= 0) {
return;
} else {
int part = partition(l, r, u[l]);
quicksort (l, part - 1);
quicksort (part + 1, r);
}
}
}

The problem is in the partition method. The pivot element is not being placed in its correct position at the end of your swaps. I've changed the method signature so that you pass in the position of the pivot element, rather than the value of the pivot, so in quicksort() you would now write:
int part = partition(l, r, l);
In the body of the pivot method, I swapped the pivot element to the end of the section (by swapping with right). So that we then ignore this element with our swaps, I took away the "+ 1" on the initialisation of rPT. I then added a statement after your while loop to move the pivot element into place. With those three changes, the method now looks like this:
public static int partition(int left, int right, int pivotPosition) {
int p = u[pivotPosition]; // pivot
// Move pivot to the end
swap(pivotPosition, right);
int lPt = left - 1;
int rPt = right;
while (true) {
while ((lPt < right) && (u[++lPt] < p));
while ((rPt > left) && (u[--rPt] > p));
if (lPt >= rPt) {
break;
} else {
swap(lPt, rPt);
System.out.println("Swapping " + lPt + " " + rPt);
}
}
// Put pivot in its place
swap(lPt, right);
return lPt;
}
With these changes, the code works for me.

You have to find a values in the left list which is larger than the pivot element and find a value in the right list which is smaller then the pivot element then we exchange the values.
package quicksort;
import java.util.Random;
import java.util.Arrays;
public class QuickSort {
/**
* #param args the command line arguments
*/
private static int[] u;
public static void main(String[] args) {
u = makeArray(10);
System.out.println(Arrays.toString(u));
quicksort(0, u.length - 1);
System.out.println(Arrays.toString(u));
}
public static int[] makeArray(int n) {
int[] a = new int[n];
int j;
Random r = new Random();
for (int i = 0; i < n; i++) {
j = (r.nextInt(100) + 1);
a[i] = j;
}
return a;
}
private static void quicksort(int low, int high) {
int i = low, j = high;
int pivot = u[low];
while (i <= j) {
while (u[i] < pivot) {
i++;
}
while (u[j] > pivot) {
j--;
}
if (i <= j) {
exchange(i, j);
i++;
j--;
}
}
if (low < j) {
quicksort(low, j); // note here
}
if (i < high) {
quicksort(i, high); // note here
}
}
private static void exchange(int i, int j) {
int temp = u[i];
u[i] = u[j];
u[j] = temp;
}
}