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# Stable QuickSort

• Difficulty Level : Easy
• Last Updated : 28 Dec, 2022

A sorting algorithm is said to be stable if it maintains the relative order of records in the case of equality of keys.

Input : (1, 5), (3, 2) (1, 2) (5, 4) (6, 4)
We need to sort key-value pairs in the increasing order of keys of first digit

There are two possible solution for the two pairs where the key is same i.e. (1, 5) and (1, 2) as shown below:
OUTPUT1: (1, 5), (1, 2), (3, 2), (5, 4), (6, 4)
OUTPUT2: (1, 2), (1, 5), (3, 2), (5, 4), (6, 4)

A stable algorithm produces first output. You can see that (1, 5) comes before (1, 2) in the sorted order, which was the original order i.e. in the given input, (1, 5) comes before (1, 2).
Second output can only be produced by an unstable algorithm. You can see that in the second output, the (1, 2) comes before (1, 5) which was not the case in the original input.

Some sorting algorithms are stable by nature like Insertion sort, Merge Sort, Bubble Sort, etc. And some sorting algorithms are not, like Heap Sort, Quick Sort, etc.
QuickSort is an unstable algorithm because we do swapping of elements according to pivot’s position (without considering their original positions).

How to make QuickSort stable?

Quicksort can be stable but it typically isn’t implemented that way. Making it stable either requires order N storage (as in a naive implementation) or a bit of extra logic for an in-place version.
In below implementation, we use extra space.

The idea is to make two separate lists:

1. First list contains items smaller than pivot.
2. Second list contains items greater than pivot.

Implementation:

## C++

 `// C++ code to implement Stable QuickSort.``// The code uses middle element as pivot.``#include ``using` `namespace` `std;` `    ``vector<``int``> quickSort(vector<``int``> ar)``    ``{``    ` `        ``// Base case``        ``if``(ar.size() <= 1)``        ``{``            ``return` `ar;``        ``}``        ``else``        ``{``        ` `            ``// Let us choose middle element a pivot           ``            ``int` `mid = ar.size() / 2;``            ``int` `pivat = ar[mid];``            ` `        ``// key element is used to break the array``            ``// into 2 halves according to their values``            ``vector<``int``> smaller ;``            ``vector<``int``> greater ;``            ` `            ``// Put greater elements in greater list,``        ``// smaller elements in smaller list. Also,``        ``// compare positions to decide where to put.       ``        ``for``(``int` `ind = 0; ind < ar.size(); ind++)``        ``{``            ``int` `val = ar[ind];``            ``if``( ind != mid )``            ``{``                ``if``( val < pivat )``                ``{``                    ``smaller.push_back(val);``                ``}``                ``else` `if``(val > pivat)``                ``{``                    ``greater.push_back(val);``                ``}``                ``else``                ``{``                    ` `                    ``// If value is same, then considering``                    ``// position to decide the list.               ``                    ``if``(ind < mid)``                    ``{``                        ``smaller.push_back(val);``                    ``}``                    ``else``                    ``{``                        ``greater.push_back(val);``                    ``}``                ``}``            ``}``        ``}``            ` `        ``vector<``int``> ans;           ``        ``vector<``int``> sa1 = quickSort(smaller);``        ``vector<``int``> sa2 = quickSort(greater);``            ` `        ``// add all elements of smaller list into ans list``        ``for``(``int` `val1 : sa1)``                ``ans.push_back(val1);``                ` `        ``// add pivat element into ans list``        ``ans.push_back(pivat);``            ` `        ``// add all elements of greater list into ans list``        ``for``(``int` `val2 : sa2)``                ``ans.push_back(val2);``            ` `        ``// return ans list``        ``return` `ans;       ``        ``}``    ``}``    ` `    ``// Driver code``    ``int` `main()``    ``{   ``        ``int` `ar[] = {10, 7, 8, 9, 1, 5};``        ``vector<``int``> al;``        ``al.push_back(10);``        ``al.push_back(7);``        ``al.push_back(8);``        ``al.push_back(9);``        ``al.push_back(1);``        ``al.push_back(5);   ``        ``vector<``int``> sortedAr = quickSort(al);``        ` `        ``for``(``int` `x:sortedAr)``        ``cout<

## Java

 `// Java code to implement Stable QuickSort.``// The code uses middle element as pivot.``import` `java.util.*;``class` `GFG``{``    ``public` `static` `ArrayList quickSort(ArrayList ar)``    ``{``      ` `        ``// Base case``        ``if``(ar.size() <= ``1``)``        ``{``            ``return` `ar;``        ``}``        ``else``        ``{``          ` `            ``// Let us choose middle element a pivot            ``            ``int` `mid = ar.size() / ``2``;``            ``int` `pivat = ar.get(mid);``            ` `           ``// key element is used to break the array``            ``// into 2 halves according to their values``            ``ArrayList smaller = ``new` `ArrayList<>();``            ``ArrayList greater = ``new` `ArrayList<>();``            ` `            ``// Put greater elements in greater list,``           ``// smaller elements in smaller list. Also,``           ``// compare positions to decide where to put.        ``           ``for``(``int` `ind = ``0``; ind < ar.size(); ind++)``           ``{``               ``int` `val = ar.get(ind);``               ``if``( ind != mid )``               ``{``                   ``if``( val < pivat )``                   ``{``                       ``smaller.add(val);``                   ``}``                   ``else` `if``(val > pivat)``                   ``{``                       ``greater.add(val);``                   ``}``                   ``else``                   ``{``                     ` `                       ``// If value is same, then considering``                       ``// position to decide the list.                   ``                       ``if``(ind < mid)``                       ``{``                           ``smaller.add(val);``                       ``}``                       ``else``                       ``{``                           ``greater.add(val);``                       ``}``                   ``}``               ``}``           ``}``           ` `           ``ArrayList ans = ``new` `ArrayList();              ``           ``ArrayList sa1 = quickSort(smaller);``           ``ArrayList sa2 = quickSort(greater);``           ` `           ``// add all elements of smaller list into ans list``           ``for``(Integer val1 : sa1)``                ``ans.add(val1);``                ` `           ``// add pivat element into ans list   ``           ``ans.add(pivat);``           ` `           ``// add all elements of greater list into ans list``           ``for``(Integer val2 : sa2)``                ``ans.add(val2);``           ` `           ``// return ans list``           ``return` `ans;        ``        ``}``    ``}``    ` `    ``// Driver code ``    ``public` `static` `void` `main(String args[])``    ``{      ``        ``int` `ar[] = {``10``, ``7``, ``8``, ``9``, ``1``, ``5``};``        ``ArrayList al = ``new` `ArrayList<>();``        ``al.add(``10``);``        ``al.add(``7``);``        ``al.add(``8``);``        ``al.add(``9``);``        ``al.add(``1``);``        ``al.add(``5``);       ``        ``ArrayList sortedAr = quickSort(al);``        ``System.out.println(sortedAr);``    ``}``}` `// This code is contributed by Naresh Saharan`

## Python3

 `# Python code to implement Stable QuickSort.``# The code uses middle element as pivot.``def` `quickSort(ar):``     ` `    ``# Base case``    ``if` `len``(ar) <``=` `1``:``        ``return` `ar` `    ``# Let us choose middle element a pivot``    ``else``:``        ``mid ``=` `len``(ar)``/``/``2``        ``pivot ``=` `ar[mid]` `        ``# key element is used to break the array``        ``# into 2 halves according to their values``        ``smaller,greater ``=` `[],[]`` ` `        ``# Put greater elements in greater list,``        ``# smaller elements in smaller list. Also,``        ``# compare positions to decide where to put.``        ``for` `indx, val ``in` `enumerate``(ar):``            ``if` `indx !``=` `mid:``                ``if` `val < pivot:``                    ``smaller.append(val)``                ``elif` `val > pivot:``                    ``greater.append(val)` `                ``# If value is same, then considering``                ``# position to decide the list.``                ``else``:``                    ``if` `indx < mid:``                        ``smaller.append(val)``                    ``else``:``                        ``greater.append(val)``        ``return` `quickSort(smaller)``+``[pivot]``+``quickSort(greater)``    ` `# Driver code to test above``ar ``=` `[``10``, ``7``, ``8``, ``9``, ``1``, ``5``]``sortedAr ``=` `quickSort(ar)``print``(sortedAr)`

## C#

 `// Include namespace system``using` `System;``using` `System.Collections.Generic;` `public` `class` `GFG``{``    ``public` `static` `List<``int``> quickSort(List<``int``> ar)``    ``{``        ``// Base case``        ``if` `(ar.Count <= 1)``        ``{``            ``return` `ar;``        ``}``        ``else``        ``{``          ` `            ``// Let us choose middle element a pivot            ``            ``var` `mid = (``int``)(ar.Count / 2);``            ``var` `pivat = ar[mid];``          ` `            ``// key element is used to break the array``            ``// into 2 halves according to their values``            ``var` `smaller = ``new` `List<``int``>();``            ``var` `greater = ``new` `List<``int``>();``          ` `            ``// Put greater elements in greater list,``            ``// smaller elements in smaller list. Also,``            ``// compare positions to decide where to put.``            ``for` `(``int` `ind = 0; ind < ar.Count; ind++)``            ``{``                ``var` `val = ar[ind];``                ``if` `(ind != mid)``                ``{``                    ``if` `(val < pivat)``                    ``{``                        ``smaller.Add(val);``                    ``}``                    ``else` `if` `(val > pivat)``                    ``{``                        ``greater.Add(val);``                    ``}``                    ``else``                    ``{``                        ``// If value is same, then considering``                        ``// position to decide the list.``                        ``if` `(ind < mid)``                        ``{``                            ``smaller.Add(val);``                        ``}``                        ``else``                        ``{``                            ``greater.Add(val);``                        ``}``                    ``}``                ``}``            ``}``            ``var` `ans = ``new` `List<``int``>();``            ``var` `sa1 = GFG.quickSort(smaller);``            ``var` `sa2 = GFG.quickSort(greater);``          ` `            ``// add all elements of smaller list into ans list``            ``foreach` `(``int` `val1 ``in` `sa1)``            ``{                ans.Add(val1);``            ``}``          ` `            ``// add pivat element into ans list   ``            ``ans.Add(pivat);``          ` `            ``// add all elements of greater list into ans list``            ``foreach` `(``int` `val2 ``in` `sa2)``            ``{                ans.Add(val2);``            ``}``          ` `            ``// return ans list``            ``return` `ans;``        ``}``    ``}``  ` `    ``// Driver code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int``[] ar = {10, 7, 8, 9, 1, 5};``        ``var` `al = ``new` `List<``int``>();``        ``al.Add(10);``        ``al.Add(7);``        ``al.Add(8);``        ``al.Add(9);``        ``al.Add(1);``        ``al.Add(5);``        ``var` `sortedAr = GFG.quickSort(al);``        ``Console.WriteLine(``string``.Join(``", "``,sortedAr));``    ``}``}` `// This code is contributed by aadityaburujwale.`

## Javascript

 `function` `quickSort(ar)``    ``{``    ` `        ``// Base case``        ``if` `(ar.length <= 1)``        ``{``            ``return` `ar;``        ``}``        ``else``        ``{``            ``// Let us choose middle element a pivot            ``            ``var` `mid = parseInt(ar.length / 2);``            ``var` `pivat = ar[mid];``            ` `            ``// key element is used to break the array``            ``// into 2 halves according to their values``            ``var` `smaller = ``new` `Array();``            ``var` `greater = ``new` `Array();``            ``var` `ind = 0;``            ` `            ``// Put greater elements in greater list,``            ``// smaller elements in smaller list. Also,``            ``// compare positions to decide where to put.``            ``for` `(ind; ind < ar.length; ind++)``            ``{``                ``var` `val = ar[ind];``                ``if` `(ind != mid)``                ``{``                    ``if` `(val < pivat)``                    ``{``                        ``(smaller.push(val) > 0);``                    ``}``                    ``else` `if` `(val > pivat)``                    ``{``                        ``(greater.push(val) > 0);``                    ``}``                    ``else``                    ``{``                        ``// If value is same, then considering``                        ``// position to decide the list.``                        ``if` `(ind < mid)``                        ``{``                            ``(smaller.push(val) > 0);``                        ``}``                        ``else``                        ``{``                            ``(greater.push(val) > 0);``                        ``}``                    ``}``                ``}``            ``}``            ``var` `ans = ``new` `Array();``            ``var` `sa1 = quickSort(smaller);``            ``var` `sa2 = quickSort(greater);``            ` `            ``// add all elements of smaller list into ans list``            ``for` `( const  val1 of sa1) {(ans.push(val1) > 0);}``            ``// add pivat element into ans list   ``            ``(ans.push(pivat) > 0);``            ` `            ``// add all elements of greater list into ans list``            ``for` `( const  val2 of sa2) {(ans.push(val2) > 0);}``            ` `            ``// return ans list``            ``return` `ans;``        ``}``    ``}``    ` `        ``var` `ar = [10, 7, 8, 9, 1, 5];``        ``var` `al = ``new` `Array();``        ``(al.push(10) > 0);``        ``(al.push(7) > 0);``        ``(al.push(8) > 0);``        ``(al.push(9) > 0);``        ``(al.push(1) > 0);``        ``(al.push(5) > 0);``        ``var` `sortedAr = quickSort(al);``        ``console.log(sortedAr);``        ` `// This code is contributed by saurabhdalal0001`

Output

`[1, 5, 7, 8, 9, 10]`

In above code, we have intentionally used middle element as a pivot to demonstrate how to consider a position as part of the comparison. Code simplifies a lot if we use last element as pivot. In the case of last element, we can always push equal elements in the smaller list.

Space Complexity:

Considering the best and average cases where the actual array will be partitioned into two sub arrays of equal sizes, i.e. the arrays smaller and greater will be of equal sizes, the recursive stack will take upto O(log n) space just like the unstable implementation of quick sort. log n will be the height of the recursion tree, at any time the maximum number of calls on the recursion stack can be O(log n). At each level of the recursion tree, extra space of “n” will be required (why? This can be easily found out by constructing the recursion tree and keeping in mind the extra space for the arrays smaller and greater). So, the total space required will be = Space used up by recursion stack + Space used up the extra arrays = O(log n) + n*O(log n) = O(nlogn). Please note that in the best and average cases, the common unstable implementation of quick sort only takes up O(log n) for the recursion stack.
In the worst case, the original array gets partitioned into two subarrays of sizes n-1 and 1. In such a case, the height of the recursion tree is “n” and at each level of recursion there will be extra space taken up by the arrays. It can be easily calculated that the worst-case space complexity comes out to be O(n2). Note that the worst-case space complexity for the common un-stable implementation is O(n).

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