I am running this script to find kth smallest element by using modified Quick Sort Algorithm. Ignore the comments they might be wrong.
I am not able to understand why do I have to (k-pivot+left-1) for right subarray. I have some idea the right subarray is shifted by pivot but I am not able to grasp it completely.
import java.util.Arrays;
public class kthMaxMin {
public static void Swap(int[] A, int id1, int id2){
int temp = A[id1];
A[id1] = A[id2];
A[id2] = temp;
}
public static int Partition(int[] A, int left, int right){
// Returns pivot index
int pivot = A[right];
int i = left; // Last position of smaller element pointer
for(int j=left;j<=right-1;j++){ // j is loop pointer
if(A[j]<=pivot){
Swap(A, i, j); // Put that element in the new position
i++;
}
}
Swap(A, i, right); // Put pivot where it belongs i.e. middle
return (i); // Location of pivot
}
public static int QuickSortModified(int[] A, int left, int right, int k){
// Returns kth smallest element
if(k>0 && k<=right-left+1){ // k is within bounds of the array
int pivot = Partition(A, left, right);
if(pivot - left == k-1){ // If number of elements in left subarray is equal to k
return A[pivot]; // then return pivot as it is kth smallest element
}
if(pivot-left > k-1){ // If size of left subarray is greater than k
return QuickSortModified(A, left, pivot-1, k); // Search in left subarray more
}
return QuickSortModified(A, pivot+1, right, k-pivot+left-1); // Else Search in right subarray
// k-pivot because right subarray is shifted by value = pivot?
}
return (-1);
}
public static void main(String[] args) {
int[] Arr = new int[10];
Arr[0] = 99;
Arr[1] = 9;
Arr[2] = 56;
Arr[3] = 78;
Arr[4] = 12;
Arr[5] = 18;
Arr[6] = 76;
Arr[7] = 19;
Arr[8] = 12;
Arr[9] = 5;
System.out.println("Array: "+Arrays.toString(Arr));
int k =5;
int min = QuickSortModified(Arr, 0, Arr.length-1, k);
System.out.println(k+"th Smallest Element: "+min);
}
}
why do I have to (k-pivot+left-1) for right subarray
Because the code is checking for pivot-left == k-1. The way the code in the question is implemented allows for finding the kth element of a portion of an array, rather than the entire array.
With an entry | helper function (QuickSelect() in this example code), the code can be simplfied:
public static int QuickSortModified(int[] A, int left, int right, int k){
int pivot = Partition(A, left, right);
if(pivot == k) // if k == pivot
return A[pivot];
if(pivot > k) // if k is in left partition
return QuickSortModified(A, left, pivot-1, k);
else // if k is in right partition
return QuickSortModified(A, pivot+1, right, k);
}
public static int QuickSelect(int[] A, int left, int right, int k){
k = k-1;
if(k < left || k > right)
return -1;
return QuickSortModified(A, left, right, k+left);
}
// ...
// calling example
// ...
int left = 2; // normally left = 0
int right = 8; // normally right = Arr.length-1
int k = 4;
int min = QuickSelect(Arr, left, right, k);
This question already has answers here:
What causes a java.lang.ArrayIndexOutOfBoundsException and how do I prevent it?
(26 answers)
Closed 4 years ago.
Im trying to make a Quicksort but it always showing up the Error ArrayIndexOutOfBoundsException.
public class Quicksort
{
void sort(int[] arr)
{
_quicksort(arr, 0, arr.length - 1);
}
private void _quicksort(int[] arr, int left, int right)
{
int pivot = (left + right)/2;
int l = left;
int r = right;
while (l <= r)
{
while (arr[l] < pivot)
{
l++;
}
while (arr[r] > pivot)
{
r--;
}
if(l <= r)
{
int temp = l;
l = r;
r = temp;
l++;
r++;
}
}
if(left < r)
{
_quicksort(arr, left, r);
}
if(l < right)
{
_quicksort(arr, l, right);
}
}
}
Does someone know why it doesnt run? It always gives a Error.
The Error message is
java.lang.ArrayIndexOutOfBoundsException: -1
at Quicksort._quicksort(Quicksort.java:18)
at Quicksort._quicksort(Quicksort.java:33)
at Quicksort.sort(Quicksort.java:5)
Error Message
It seems like there are a couple of issues with your code. I've listed them below:
The variable pivot stores the index of the pivot element and not the actual value. So, you can't use pivot for comaparison as you have done in the 2 nested while loops. You need arr[pivot] instead of pivot there.
Imagine arr looks like {1, 1, 1, 3, 2, 2, 2}. Here, pivot will be equal to 3 i.e. arr[pivot] will be equal to 3. Now, after both the nested while loops terminate, l will be equal to 3 and r will remain equal to 6. You then swap arr[l] and arr[r] and increment both l and r. Since l is still less than equal to r, the outer while loop runs for a second time and you'll get an ArrayIndexOutOfBoundsExecption when the control reaches the second nested while loop. This happens because you're trying to access arr[7] (Out of Bounds).
Here's my code:
class Quicksort
{
void sort(int[] arr)
{
myQuicksort(arr, 0, arr.length - 1);
}
private void myQuicksort(int[] arr, int l, int r) {
if (l >= r) {
return;
}
int pivotIndex = (l + r) / 2;
swap (arr, r, pivotIndex);
int pivotValue = arr[r];
int swapIndex = 0;
int currentIndex = 0;
while (currentIndex != r) {
if (arr[currentIndex] < pivotValue) {
swap(arr, currentIndex, swapIndex);
swapIndex++;
}
currentIndex++;
}
swap(arr, r, swapIndex);
myQuicksort(arr, l, swapIndex - 1);
myQuicksort(arr, swapIndex + 1, r);
}
private void swap(int[] arr, int i, int j) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}
public class Main{
public static void main(String[] args) {
Quicksort quicksort = new Quicksort();
int[] arr = {3, 7, 1, 0, 4};
quicksort.sort(arr);
for (int i : arr) {
System.out.println(i);
}
}
}
You should read up on Quicksort. But here are the main points:
Choose a random pivot element and swap it with the last element. This makes the implementation much simpler.
Loop over the input array and keep a track of a swapIndex such that everything before the swapIndex is less than the pivotValue and everything from the swapIndex till the currentIndex is greater than (or equal) the pivotValue.
After the loop runs out, swap the element at swapIndex with the pivot. This inserts the pivot in its correct position.
The pivot divides the array into 2 subarrays. Call myQuicksort on these 2 subarrays.
I'm running this and I am being told it would not run fast enough. What is a good way to increase the speed of this running class? I am guessing I would need to change my nested while loops. That is the only thing I can think of. The if statements should all be linear...
import java.io.File;
import java.io.FileNotFoundException;
import java.util.*;
public class QSortLab {
static int findpivot(Comparable[] A, int i, int j) {
return (i + j) / 2;
}
static <E> void swap(E[] A, int p1, int p2) {
E temp = A[p1];
A[p1] = A[p2];
A[p2] = temp;
}
static void quicksort(Comparable[] A, int i, int j) { // Quicksort
int pivotindex = findpivot(A, i, j); // Pick a pivot
swap(A, pivotindex, j); // Stick pivot at end
int k = partition(A, i, j-1, A[j]);
swap(A, k, j); // Put pivot in place
if ((k-i) > 1) quicksort(A, i, k-1); // Sort left partition
if ((j-k) > 1) quicksort(A, k+1, j); // Sort right partition
}
static int partition(Comparable[] A, int left, int right, Comparable pivot) {
while (left <= right) { // Move bounds inward until they meet
while (A[left].compareTo(pivot) < 0) left++;
while ((right >= left) && (A[right].compareTo(pivot) >= 0)) right--;
if (right > left) swap(A, left, right); // Swap out-of-place values
}
return left; // Return first position in right partition
}
}
What do you mean you need to change your nested while loops? Quick Sort is defined by those features. Removing wouldn't function properly.
As for optimization, by default it should be known that primitives vs objects tend to be different. E.g. primitives on stack/heap to keep stack small & heap stores object with refs able to be on stack.
So let's test some stuff
primitive quick sort (from here)
Integer quick sort (same code as above, but with Integer class)
Your original posted code
Your original posted code (w/ several edits)
Here's the entire code I used.
import java.util.Random;
public class App {
public static final int ARR_SIZE = 1000;
public static final int TEST_ITERS = 10000;
public static Random RANDOM = new Random();
public static void main(String[] args) {
int[] a = new int[ARR_SIZE];
Integer[] b = new Integer[ARR_SIZE];
Integer[] c = new Integer[ARR_SIZE];
Integer[] d = new Integer[ARR_SIZE];
long sum = 0, start = 0, end = 0;
for (int i = 0; i < TEST_ITERS; ++i) {
for (int j = 0; j < ARR_SIZE; ++j)
a[j] = RANDOM.nextInt();
start = System.nanoTime();
quickSort(a, 0, a.length - 1);
end = System.nanoTime();
sum += (end - start);
}
System.out.println((sum / TEST_ITERS) + " nano, qs avg - 'int'");
sum = 0;
for (int i = 0; i < TEST_ITERS; ++i) {
for (int j = 0; j < ARR_SIZE; ++j)
b[j] = RANDOM.nextInt();
start = System.nanoTime();
quickSort(b, 0, b.length - 1);
end = System.nanoTime();
sum += (end - start);
}
System.out.println((sum / TEST_ITERS) + " nano, qs avg - 'Integer'");
sum = 0;
for (int i = 0; i < TEST_ITERS; ++i) {
for (int j = 0; j < ARR_SIZE; ++j)
c[j] = RANDOM.nextInt();
start = System.nanoTime();
quicksort(c, 0, c.length - 1);
end = System.nanoTime();
sum += (end - start);
}
System.out.println((sum / TEST_ITERS) + " nano, qs avg - 'Comparable' (SO user code)");
sum = 0;
for (int i = 0; i < TEST_ITERS; ++i) {
for (int j = 0; j < ARR_SIZE; ++j)
d[j] = RANDOM.nextInt();
start = System.nanoTime();
qs_quicksort(d, 0, d.length - 1);
end = System.nanoTime();
sum += (end - start);
}
System.out.println((sum / TEST_ITERS) + " nano, qs avg - 'Comparable' (SO user code - edit)");
for (int i = 0; i < ARR_SIZE; ++i) {
final int n = RANDOM.nextInt();
a[i] = n;
b[i] = n;
c[i] = n;
d[i] = n;
}
quickSort(a, 0, a.length - 1);
Integer[] aConv = new Integer[ARR_SIZE];
for (int i = 0; i < ARR_SIZE; ++i)
aConv[i] = a[i];
quickSort(b, 0, b.length - 1);
quicksort(c, 0, c.length - 1);
qs_quicksort(d, 0, d.length - 1);
isSorted(new Integer[][] { aConv, b, c, d });
System.out.println("All properly sorted");
}
public static void isSorted(Integer[][] arrays) {
if (arrays.length != 4) {
System.out.println("error sorting, input arr len");
return;
}
for (int i = 0; i < ARR_SIZE; ++i) {
int val1 = arrays[0][i].compareTo(arrays[1][i]);
int val2 = arrays[1][i].compareTo(arrays[2][i]);
int val3 = arrays[2][i].compareTo(arrays[3][i]);
if (val1 != 0 || val2 != 0 || val3 != 00) {
System.out.printf("Error [i = %d]: a = %d, b = %d, c = %d", i, arrays[0][i], arrays[1][i], arrays[2][i], arrays[3][i]);
break;
}
}
}
public static int partition(int arr[], int left, int right) {
int i = left, j = right;
int tmp;
int pivot = arr[(left + right) / 2];
while (i <= j) {
while (arr[i] < pivot)
i++;
while (arr[j] > pivot)
j--;
if (i <= j) {
tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
i++;
j--;
}
}
return i;
}
public static void quickSort(int arr[], int left, int right) {
int index = partition(arr, left, right);
if (left < index - 1)
quickSort(arr, left, index - 1);
if (index < right)
quickSort(arr, index, right);
}
public static int partition(Integer[] arr, int left, int right) {
int i = left, j = right;
Integer pivot = arr[(left + right) / 2];
while (i <= j) {
while (arr[i].compareTo(pivot) < 0)
i++;
while (arr[j].compareTo(pivot) > 0)
j--;
if (i <= j) {
Integer temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
i++;
j--;
}
}
return i;
}
public static void quickSort(Integer[] arr, int left, int right) {
int index = partition(arr, left, right);
if (left < index - 1)
quickSort(arr, left, index - 1);
if (index < right)
quickSort(arr, index, right);
}
static int findpivot(Comparable[] A, int i, int j)
{
return (i+j)/2;
}
static <E> void swap(E[] A, int p1, int p2) {
E temp = A[p1];
A[p1] = A[p2];
A[p2] = temp;
}
static void quicksort(Comparable[] A, int i, int j) { // Quicksort
int pivotindex = findpivot(A, i, j); // Pick a pivot
swap(A, pivotindex, j); // Stick pivot at end
int k = partition(A, i, j-1, A[j]);
swap(A, k, j); // Put pivot in place
if ((k-i) > 1) quicksort(A, i, k-1); // Sort left partition
if ((j-k) > 1) quicksort(A, k+1, j); // Sort right partition
}
static int partition(Comparable[] A, int left, int right, Comparable pivot) {
while (left <= right) { // Move bounds inward until they meet
while (A[left].compareTo(pivot) < 0) left++;
while ((right >= left) && (A[right].compareTo(pivot) >= 0)) right--;
if (right > left) swap(A, left, right); // Swap out-of-place values
}
return left; // Return first position in right partition
}
static <E> void qs_swap(E[] A, int p1, int p2) {
E temp = A[p1];
A[p1] = A[p2];
A[p2] = temp;
}
static void qs_quicksort(Comparable[] A, int i, int j) { // Quicksort
int pivotindex = (i+j)/2;
qs_swap(A, pivotindex, j); // Stick pivot at end
int k = qs_partition(A, i, j-1, A[j]);
qs_swap(A, k, j); // Put pivot in place
if ((k-i) > 1) qs_quicksort(A, i, k-1); // Sort left partition
if ((j-k) > 1) qs_quicksort(A, k+1, j); // Sort right partition
}
static int qs_partition(Comparable[] A, int left, int right, Comparable pivot) {
while (left <= right) { // Move bounds inward until they meet
while (A[left].compareTo(pivot) < 0) left++;
while ((right >= left) && (A[right].compareTo(pivot) >= 0)) right--;
if (right > left) { qs_swap(A, left, right); // Swap out-of-place values
left++; right--;}
}
return left; // Return first position in right partition
}
}
This produces the output:
56910 nano, qs avg - 'int'
69498 nano, qs avg - 'Integer'
76762 nano, qs avg - 'Comparable' (SO user code)
71846 nano, qs avg - 'Comparable' (SO user code - edit)
All properly sorted
Now, breaking down the results
The 'int' vs 'Integer' shows great diff when simply using primitives vs non-primitives (I'm sure at some points in the code there may be boxing but hopefully not in critical spots ;) - please edit this if so). The 'int' vs 'Integer' uses same code with exception of 'int' 'Integer'. See the following four method signatures that are used in this comparison, 'int'
public static int partition(int arr[], int left, int right)
public static void quickSort(int arr[], int left, int right)
and 'Integer'
public static int partition(Integer[] arr, int left, int right)
public static void quickSort(Integer[] arr, int left, int right)
respectively.
Then there are the method signatures related to the original code you posted,
static int findpivot(Comparable[] A, int i, int j)
static <E> void swap(E[] A, int p1, int p2)
static void quicksort(Comparable[] A, int i, int j)
static int partition(Comparable[] A, int left, int right, Comparable pivot)
and the modified ones,
static <E> void qs_swap(E[] A, int p1, int p2)
static void qs_quicksort(Comparable[] A, int i, int j)
static int qs_partition(Comparable[] A, int left, int right, Comparable pivot)
As you can see, in the modified code, findpivot was removed directly and replaced into the calling spot in quicksort. Also, the partition method gained counters for left and right respectively. left++; right--;
And finally, to ensure these 4 variations of quicksort actually did the sole purpose, sort, I added a method, isSorted() to check the validity of the same generated content and that it's sorted accordingly based on each of the 4 different sorts.
In conclusion, I think my edits may have saved a portion of time/nanoseconds, however I wasn't able to achieve the same time as the Integer test. Hopefully I've not missed anything obvious and edits are welcome if need be. Cheers
Well, I couldn't tell from testing whether this makes any difference at all because the timer on my machine is terrible , but I think most of the work in this algo is done with the swap function, so thinking about how to make that in particular more efficient, maybe the function call/return itself consumes cycles, and perhaps the creation of the temp variable each time the function is called also takes cycles, so maybe the code would be more efficient if the swap work was done in line. It was not obvious though when I tested on my machine as the nanotimer returned results +/- 20% each time I ran the program
public class QSort2 {
static int findpivot(Comparable[] A, int i, int j) {
return (i + j) / 2;
}
static Comparable temp;
static void quicksort(Comparable[] A, int i, int j) { // Quicksort
int pivotindex = findpivot(A, i, j); // Pick a pivot
// swap(A, pivotindex, j); // Stick pivot at end
temp = A[pivotindex];
A[pivotindex] = A[j];
A[j] = temp;
int k = partition(A, i, j - 1, A[j]);
//swap(A, k, j); // Put pivot in place
temp = A[k];
A[k] = A[j];
A[j] = temp;
if ((k - i) > 1) quicksort(A, i, k - 1); // Sort left partition
if ((j - k) > 1) quicksort(A, k + 1, j); // Sort right partition
}
static int partition(Comparable[] A, int left, int right, Comparable pivot) {
while (left <= right) { // Move bounds inward until they meet
while (A[left].compareTo(pivot) < 0) left++;
while ((right >= left) && (A[right].compareTo(pivot) >= 0)) right--;
if (right > left) {
//swap(A, left, right);} // Swap out-of-place values
temp = A[left];
A[left] = A[right];
A[right] = temp;
}
}
return left; // Return first position in right partition
}
}
Firstly (as the question title implies) I'm not looking for why the bellow partitioning method doesn't work, rather a modification to it so that it will work for the following input:
int[] a = {1,2,8,4,5,7};
Here's the partition method along with some other stuff:
static int[] swap (int[] a,int i,int j){
int t = a[i];
a[i] = a[j];
a[j] = t;
return a;
}
static int partition (int[] a,int l,int r){
int i = l;
int j = r;
int v = a[l];
while (true) {
while (a[i] < v) {
if (i == r) break;
i++;
}
while (a[j] > v) {
if (j == l) break;
j--;
}
if (i >= j) break;
a = swap(a,i,j);
}
a = swap(a, l, j);
return j;
}
void sort(int[] a,int l,int r){
int j = partition(a, l, r);
sort(a, l, j-1);
sort(a, j+1, r);
}
public static void main(String[] args) {
int[] a = {1,2,8,4,5,7};
System.out.println(partition(a,0,5));
}
Output:
0
The output is the index of the pivot returned from the partition method. 0, as the index of the pivot, makes sense in terms of the definition, i.e. everything left of the pivot is smaller and everything right of the pivot is larger, but clearly runs into a problem in sort namely:
sort(a, l, j-1);
where you have the right pointer being negative (j-1 = 0-1 = -1). My question again being is there a modification to the above method(s) that will maintain the definition (everything left of the pivot is smaller and everything right of the pivot is larger) and not run into the problem in sort.
The missing part is the line
if ( l >= r ) return;
in the beginning of the sort method. This is actually the recursion stop step so it is necessary to have it anyway to prevent endless recursion. But besides that, it also solves your problem, because if you call sort(0,-1) then -1 is less than 0, so it prevents further processing of that index.
I have two implementation of quick sort the first one uses a median of (fist ,middle , last ) as pivot and the second uses the middle element as pivot
the first Implementation :
public class quickMedian {
public void sort(int array[])
// pre: array is full, all elements are non-null integers
// post: the array is sorted in ascending order
{
quickSort(array, 0, array.length - 1); // quicksort all the elements in the array
}
public void quickSort(int array[], int start, int end)
{
int i = start; // index of left-to-right scan
int k = end; // index of right-to-left scan
if (end - start >= 1) // check that there are at least two elements to sort
{
if (array[start+(end-start)/2]>array[end]){
swap(array,start+(end-start)/2, end);
}
if (array[start]>array[end]){
swap(array,start, end);
}
if (array[start+(end-start)/2]>array[start]){
swap(array, start+(end-start)/2, start);
}
int pivot = array[start]; // set the pivot as the first element in the partition
while (k > i) // while the scan indices from left and right have not met,
{
while (array[i] <= pivot && i <= end && k > i) // from the left, look for the first
i++; // element greater than the pivot
while (array[k] > pivot && k >= start && k >= i) // from the right, look for the first
k--; // element not greater than the pivot
if (k > i) // if the left seekindex is still smaller than
swap(array, i, k); // the right index, swap the corresponding elements
}
swap(array, start, k); // after the indices have crossed, swap the last element in // the left partition with the pivot
quickSort(array, start, k - 1); // quicksort the left partition
quickSort(array, k + 1, end); // quicksort the right partition
}
else // if there is only one element in the partition, do not do any sorting
{
return; // the array is sorted, so exit
}
}
public void swap(int array[], int index1, int index2)
// pre: array is full and index1, index2 < array.length
// post: the values at indices 1 and 2 have been swapped
{
int temp = array[index1]; // store the first value in a temp
array[index1] = array[index2]; // copy the value of the second into the first
array[index2] = temp; // copy the value of the temp into the second
}
}
The second Implementation :
public class quickSort {
private int array[];
private int length;
public void sort(int[] inputArr) {
if (inputArr == null || inputArr.length == 0) {
return;
}
this.array = inputArr;
length = inputArr.length;
quickSorter(0, length - 1);
}
private void quickSorter(int lowerIndex, int higherIndex) {
int i = lowerIndex;
int j = higherIndex;
// calculate pivot number, I am taking pivot as middle index number
int pivot = array[lowerIndex+(higherIndex-lowerIndex)/2];
// Divide into two arrays
while (i <= j) {
/**
* In each iteration, we will identify a number from left side which
* is greater then the pivot value, and also we will identify a number
* from right side which is less then the pivot value. Once the search
* is done, then we exchange both numbers.
*/
while (array[i] < pivot) {
i++;
}
while (array[j] > pivot) {
j--;
}
if (i <= j) {
exchangeNumbers(i, j);
//move index to next position on both sides
i++;
j--;
}
}
// call quickSort() method recursively
if (lowerIndex < j)
quickSorter(lowerIndex, j);
if (i < higherIndex)
quickSorter(i, higherIndex);
}
private void exchangeNumbers(int i, int j) {
int temp = array[i];
array[i] = array[j];
array[j] = temp;
}
}
To obtain the median we need to do extra steps on each recursion which may increase the time a little bit (if the array is totally random) .
I am testing these two classes on an array of size N=10,000,000 randomly generated and I have done the test many times the first Implementation takes around 30 seconds and the second takes around 4 seconds
so this is obviously not caused by the extra overhead to get the median of three numbers .
There must be something wrong with the first implementation, what is it ?
here is the testing code :
public static void main(String[] args) {
File number = new File ("f.txt");
final int size = 10000000;
try{
// quickSort s = new quickSort();
quickMedian s = new quickMedian();
writeTofile(number, size);
int [] arr1 =readFromFile(number, size);
long a=System.currentTimeMillis();
s.sort(arr1);
long b=System.currentTimeMillis();
System.out.println("quickSort: "+(double)(b-a)/1000);
}catch (Exception ex){ex.printStackTrace();}
}