Odd-Even Sort / Brick Sort
This is basically a variation of bubble-sort. This algorithm is divided into two phases- Odd and Even Phase. The algorithm runs until the array elements are sorted and in each iteration two phases occurs- Odd and Even Phases.
In the odd phase, we perform a bubble sort on odd indexed elements and in the even phase, we perform a bubble sort on even indexed elements.
C++
#include<bits/stdc++.h>
using namespace std;
void oddEvenSort( int arr[], int n)
{
bool isSorted = false ;
while (!isSorted)
{
isSorted = true ;
for ( int i=1; i<=n-2; i=i+2)
{
if (arr[i] > arr[i+1])
{
swap(arr[i], arr[i+1]);
isSorted = false ;
}
}
for ( int i=0; i<=n-2; i=i+2)
{
if (arr[i] > arr[i+1])
{
swap(arr[i], arr[i+1]);
isSorted = false ;
}
}
}
return ;
}
void printArray( int arr[], int n)
{
for ( int i=0; i < n; i++)
cout << arr[i] << " " ;
cout << "\n" ;
}
int main()
{
int arr[] = {34, 2, 10, -9};
int n = sizeof (arr)/ sizeof (arr[0]);
oddEvenSort(arr, n);
printArray(arr, n);
return (0);
}
|
Java
import java.io.*;
class GFG
{
public static void oddEvenSort( int arr[], int n)
{
boolean isSorted = false ;
while (!isSorted)
{
isSorted = true ;
int temp = 0 ;
for ( int i= 1 ; i<=n- 2 ; i=i+ 2 )
{
if (arr[i] > arr[i+ 1 ])
{
temp = arr[i];
arr[i] = arr[i+ 1 ];
arr[i+ 1 ] = temp;
isSorted = false ;
}
}
for ( int i= 0 ; i<=n- 2 ; i=i+ 2 )
{
if (arr[i] > arr[i+ 1 ])
{
temp = arr[i];
arr[i] = arr[i+ 1 ];
arr[i+ 1 ] = temp;
isSorted = false ;
}
}
}
return ;
}
public static void main (String[] args)
{
int arr[] = { 34 , 2 , 10 , - 9 };
int n = arr.length;
oddEvenSort(arr, n);
for ( int i= 0 ; i < n; i++)
System.out.print(arr[i] + " " );
System.out.println( " " );
}
}
|
Python3
def oddEvenSort(arr, n):
isSorted = 0
while isSorted = = 0 :
isSorted = 1
temp = 0
for i in range ( 1 , n - 1 , 2 ):
if arr[i] > arr[i + 1 ]:
arr[i], arr[i + 1 ] = arr[i + 1 ], arr[i]
isSorted = 0
for i in range ( 0 , n - 1 , 2 ):
if arr[i] > arr[i + 1 ]:
arr[i], arr[i + 1 ] = arr[i + 1 ], arr[i]
isSorted = 0
return
arr = [ 34 , 2 , 10 , - 9 ]
n = len (arr)
oddEvenSort(arr, n);
for i in range ( 0 , n):
print (arr[i], end = ' ' )
|
C#
using System;
class GFG
{
public static void oddEvenSort( int []arr, int n)
{
bool isSorted = false ;
while (!isSorted)
{
isSorted = true ;
int temp =0;
for ( int i = 1; i <= n - 2; i = i + 2)
{
if (arr[i] > arr[i+1])
{
temp = arr[i];
arr[i] = arr[i+1];
arr[i+1] = temp;
isSorted = false ;
}
}
for ( int i = 0; i <= n - 2; i = i + 2)
{
if (arr[i] > arr[i+1])
{
temp = arr[i];
arr[i] = arr[i+1];
arr[i+1] = temp;
isSorted = false ;
}
}
}
return ;
}
public static void Main ()
{
int []arr = {34, 2, 10, -9};
int n = arr.Length;
oddEvenSort(arr, n);
for ( int i = 0; i < n; i++)
Console.Write(arr[i] + " " );
Console.WriteLine( " " );
}
}
|
Javascript
<script>
function oddEvenSort(arr, n)
{
let isSorted = false ;
while (!isSorted)
{
isSorted = true ;
let temp =0;
for (let i=1; i<=n-2; i=i+2)
{
if (arr[i] > arr[i+1])
{
temp = arr[i];
arr[i] = arr[i+1];
arr[i+1] = temp;
isSorted = false ;
}
}
for (let i=0; i<=n-2; i=i+2)
{
if (arr[i] > arr[i+1])
{
temp = arr[i];
arr[i] = arr[i+1];
arr[i+1] = temp;
isSorted = false ;
}
}
}
return ;
}
let arr = [34, 2, 10, -9];
let n = arr.length;
oddEvenSort(arr, n);
for (let i=0; i < n; i++)
document.write(arr[i] + " " );
document.write( " " );
</script>
|
Output :
-9 2 10 34
We demonstrate the above algorithm using the below illustration on the array = {3, 2, 3, 8, 5, 6, 4, 1}
Please refer wiki for proof of correctness.
Time Complexity : O(N2) where, N = Number of elements in the input array.
Auxiliary Space : O(1). Just like bubble sort this is also an in-place algorithm.
Exercise
In our program in each iteration we first do bubble sort on odd indexed elements and then a bubble sort on the even indexed elements.
Will we get a sorted result if we first perform a bubble sort on even indexed element first and then on the odd indexed element ?
References
https://en.wikipedia.org/wiki/Odd%E2%80%93even_sort
Last Updated :
31 Jan, 2023
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