Stable QuickSort
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:
- First list contains items smaller than pivot.
- Second list contains items greater than pivot.
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
vector< int > quickSort(vector< int > ar)
{
if (ar.size() <= 1)
{
return ar;
}
else
{
int mid = ar.size() / 2;
int pivat = ar[mid];
vector< int > smaller ;
vector< int > greater ;
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 (ind < mid)
{
smaller.push_back(val);
}
else
{
greater.push_back(val);
}
}
}
}
vector< int > ans;
vector< int > sa1 = quickSort(smaller);
vector< int > sa2 = quickSort(greater);
for ( int val1 : sa1)
ans.push_back(val1);
ans.push_back(pivat);
for ( int val2 : sa2)
ans.push_back(val2);
return ans;
}
}
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<<x<< " " ;
}
|
Java
import java.util.*;
class GFG
{
public static ArrayList<Integer> quickSort(ArrayList<Integer> ar)
{
if (ar.size() <= 1 )
{
return ar;
}
else
{
int mid = ar.size() / 2 ;
int pivat = ar.get(mid);
ArrayList<Integer> smaller = new ArrayList<>();
ArrayList<Integer> greater = new ArrayList<>();
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 (ind < mid)
{
smaller.add(val);
}
else
{
greater.add(val);
}
}
}
}
ArrayList<Integer> ans = new ArrayList<Integer>();
ArrayList<Integer> sa1 = quickSort(smaller);
ArrayList<Integer> sa2 = quickSort(greater);
for (Integer val1 : sa1)
ans.add(val1);
ans.add(pivat);
for (Integer val2 : sa2)
ans.add(val2);
return ans;
}
}
public static void main(String args[])
{
int ar[] = { 10 , 7 , 8 , 9 , 1 , 5 };
ArrayList<Integer> al = new ArrayList<>();
al.add( 10 );
al.add( 7 );
al.add( 8 );
al.add( 9 );
al.add( 1 );
al.add( 5 );
ArrayList<Integer> sortedAr = quickSort(al);
System.out.println(sortedAr);
}
}
|
Python3
def quickSort(ar):
if len (ar) < = 1 :
return ar
else :
mid = len (ar) / / 2
pivot = ar[mid]
smaller,greater = [],[]
for indx, val in enumerate (ar):
if indx ! = mid:
if val < pivot:
smaller.append(val)
elif val > pivot:
greater.append(val)
else :
if indx < mid:
smaller.append(val)
else :
greater.append(val)
return quickSort(smaller) + [pivot] + quickSort(greater)
ar = [ 10 , 7 , 8 , 9 , 1 , 5 ]
sortedAr = quickSort(ar)
print (sortedAr)
|
C#
using System;
using System.Collections.Generic;
public class GFG
{
public static List< int > quickSort(List< int > ar)
{
if (ar.Count <= 1)
{
return ar;
}
else
{
var mid = ( int )(ar.Count / 2);
var pivat = ar[mid];
var smaller = new List< int >();
var greater = new List< int >();
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 (ind < mid)
{
smaller.Add(val);
}
else
{
greater.Add(val);
}
}
}
}
var ans = new List< int >();
var sa1 = GFG.quickSort(smaller);
var sa2 = GFG.quickSort(greater);
foreach ( int val1 in sa1)
{ ans.Add(val1);
}
ans.Add(pivat);
foreach ( int val2 in sa2)
{ ans.Add(val2);
}
return ans;
}
}
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));
}
}
|
Javascript
function quickSort(ar)
{
if (ar.length <= 1)
{
return ar;
}
else
{
var mid = parseInt(ar.length / 2);
var pivat = ar[mid];
var smaller = new Array();
var greater = new Array();
var ind = 0;
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 (ind < mid)
{
(smaller.push(val) > 0);
}
else
{
(greater.push(val) > 0);
}
}
}
}
var ans = new Array();
var sa1 = quickSort(smaller);
var sa2 = quickSort(greater);
for ( const val1 of sa1) {(ans.push(val1) > 0);}
(ans.push(pivat) > 0);
for ( const val2 of sa2) {(ans.push(val2) > 0);}
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);
|
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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