# Quicksort Array in Java

Quicksort is a divide and conquer algorithm. It first divides a large list into two smaller sub-lists and then recursively sort the two sub-lists. If we want to sort an array without any extra space, quicksort is a good option. On average, time complexity is O(n log(n)).

The basic step of sorting an array are as follows:

• Select a pivot
• Move smaller elements to the left and move bigger elements to the right of the pivot
• Recursively sort left part and right part

This post shows two versions of the Java implementation. The first one picks the rightmost element as the pivot and the second one picks the middle element as the pivot.

Version 1: Rightmost element as pivot

The following is the Java Implementation using rightmost element as the pivot.

```public class QuickSort {   public static void main(String[] args) { int[] arr = {4, 5, 1, 2, 3, 3}; quickSort(arr, 0, arr.length-1); System.out.println(Arrays.toString(arr)); }   public static void quickSort(int[] arr, int start, int end){   int partition = partition(arr, start, end);   if(partition-1>start) { quickSort(arr, start, partition - 1); } if(partition+1<end) { quickSort(arr, partition + 1, end); } }   public static int partition(int[] arr, int start, int end){ int pivot = arr[end];   for(int i=start; i<end; i++){ if(arr[i]<pivot){ int temp= arr[start]; arr[start]=arr[i]; arr[i]=temp; start++; } }   int temp = arr[start]; arr[start] = pivot; arr[end] = temp;   return start; } }```

You can use the example below to go through the code.

Version 2: Middle element as pivot

```public class QuickSort { public static void main(String[] args) { int[] x = { 9, 2, 4, 7, 3, 7, 10 }; System.out.println(Arrays.toString(x));   int low = 0; int high = x.length - 1;   quickSort(x, low, high); System.out.println(Arrays.toString(x)); }   public static void quickSort(int[] arr, int low, int high) { if (arr == null || arr.length == 0) return;   if (low >= high) return;   // pick the pivot int middle = low + (high - low) / 2; int pivot = arr[middle];   // make left < pivot and right > pivot int i = low, j = high; while (i <= j) { while (arr[i] < pivot) { i++; }   while (arr[j] > pivot) { j--; }   if (i <= j) { int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; i++; j--; } }   // recursively sort two sub parts if (low < j) quickSort(arr, low, j);   if (high > i) quickSort(arr, i, high); } }```

Output:

9 2 4 7 3 7 10
2 3 4 7 7 9 10

Here is a very good animation of quicksort.

Category >> Algorithms
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• Tasneem

With respect to Version 1: To avoid unnecessary swap, in the function partition we can add the below check :
int temp = arr[start];
if(temp != pivot){
arr[start] = pivot;
arr[end] = temp;
}

• Taimoor

The code given up produces a StackOverflow Error, and it does not compile. Can someone suggest why that might be the reason? If you change the test array, the correct output is not shown.

• Great explanation!! Understood the concept. Thanks for sharing.

• mj

Mathematically it is correct,but it is used to avoid int overflow.(int has an upper limit of 65535)

• Chris

@Sorter

if (arr == null || arr.length == 0)
return;
i would write in main:
if( x.length != 0 )

This would be a null pointer exception if x is null, hence a null check is required. As a stand-alone program it doesn’t matter, because you already know the inputs are good, but it’s good practice to handle it anyway.

But I would actually use arr.length > 1 is even faster

This is a misconception. Perhaps this was true decades ago, but the compiler knows that a division by 2 is equivalent to a bit shift by 1. Same with multiplication. It’s easy to do a benchmark, or just examine the Java byte code and you’ll see it’s the same.

AKA, don’t do this, because it may be confusing for developers who come behind you who may not recognize that the bit shift is division/multiplication.

@kk
I could be wrong but I think the running time for the partition part could be O(n^2) so this Quicksort does not have O(nlgn) running time.

This is sort of true. The worst case runtime of quicksort is n^2, but on average it is nlgn, so we say it is nlgn.

• sorter2

int middle = low + (high – low) / 2;
int pivot = arr[middle];
i would write:
int pivot = arr[ (high + low ) / 2];

Because i have knowledge in math, i can make your math expression less complicate for you:
low + (high – low) / 2
= low + high/2 – low/2
= ( low – low/2 ) + high/2
= low/2 + high/2
= (low + high) / 2

• sorter

int middle = low + (high – low) / 2;
int pivot = arr[middle];
i would write:
int pivot = arr[ (high + low ) / 2];

Because i have knowledge in math, i can make your math expression less complicate for you:
low + (high – low) / 2
= low + high/2 – low/2
= ( low – low/2 ) + high/2
= low/2 + high/2
= (low + high) / 2

• sorter

if (arr == null || arr.length == 0)
return;
i would write in main:
if( x.length != 0 )
{
quickSort(x, low, high);
System.out.println(Arrays.toString(x));
}

Running time is N log N for the average input.
Each recursion runs log N (average) times, and each run is bounded to N

• Christopher Angulo Avila

Hmm, I feel bad for not being able to implement this myself ,,,, 🙁

• Jack

Doesn’t one of the pivot conditionals need to have an equals, such as =, instead of ? Otherwise, what happens if the pivot value occurs multiple times in the array?

• Zinga Zee

try this

public void quickSort(int[] arr, int p, int r){

if(p < r) {

int q = partition(arr, p, r);

quickSort(arr, p, q-1);

quickSort(arr, q+1, r);

}

}

public int partition(int arr[], int p, int r)

{

int i = p-1;

int pivot = arr[r];

for (int j = p; j <= r; j++) {

if(arr[j] <= pivot){

i++;

//do the swap

if(i!=j){

arr[i] = arr[i] ^ arr[j];

arr[j] = arr[i] ^ arr[j];

arr[i] = arr[i] ^ arr[j];

}

}

}

return i;

}

• Nate Neu

Just a real hint ( low + high ) >> 1 is even faster

• Sergey Dinamik

I’m thinking about it 30 mins already ) voting for “yes, we can”

• kk

I could be wrong but I think the running time for the partition part could be O(n^2) so this Quicksort does not have O(nlgn) running time.

Can we remove the if conditions here :

``` if (low i) quickSort(arr, i, high); ```

Inside the function we are again checking the same.

• Nanda firizki

i like this program and i understand it , thanks bro

• Dylan Yiyang Qiu

In recursively sort two sub parts, it seems you don’t need to check relationship between low and j (high and i) before you call quickSort. Since you define a stop condition of low>=high at the beginning.

• u2

2(low)/2+(high-low)/2 = (2low-low+high)/2 = (low+high)/2.

• Elver

shouldn’t “if(i <= j)" just be "if(i < j)"? What is the point to swap i and j if both index where equal?

• JavaPrograms

check this program with explanation http://goo.gl/6d529h

• theLastUnicorn

ah no? He’s referring to get the difference first: (high – low), then use this divided by 2, then add to the starting position, which “low+” will be the last operation.

• crackerplace

I just wanted to understand one point.The above quickSort method excluding the recursion part modifies the array such that all elements less than pivot are on left side and all elements greater than pivot are on right side.Is this correct.My point is that the pivot will not be at the boundary but it will be somewhere in the right part ?

• Pravesh Jain

Just a little problem with your statement. You mean to say they will not retain their *order* after sorting. Not places.

• Govind

Good tutorial, but it’s important to note that it’s not stable. If you have same numbers they will not retain their places after sorting. By the way you can also use following quicksort algorithm to sort it in-place.

• Ashish Thakran

Quicksort is slightly sensitive to input that happens to be in the right order, in which case it can skip some swaps. Mergesort doesn’t have any such optimizations, which also makes Quicksort a bit faster compared to Mergesort.

To know more about quicksort and mergesort, below link can be useful

Why Quick sort is better than Merge sort

• kd

Just a little hint: (low + high) / 2 = low + (high – low) / 2.