# 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

```// A C++ Program to implement Odd-Even / Brick Sort
#include<bits/stdc++.h>
using namespace std;

// A function to sort the algorithm using Odd Even sort
void oddEvenSort(int arr[], int n)
{
bool isSorted = false; // Initially array is unsorted

while (!isSorted)
{
isSorted = true;

// Perform Bubble sort on odd indexed element
for (int i=1; i<=n-2; i=i+2)
{
if (arr[i] > arr[i+1])
{
swap(arr[i], arr[i+1]);
isSorted = false;
}
}

// Perform Bubble sort on even indexed element
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;
}

// A utility function ot print an array of size n
void printArray(int arr[], int n)
{
for (int i=0; i < n; i++)
cout << arr[i] << " ";
cout << "\n";
}

// Driver program to test above functions.
int main()
{
int arr[] = {34, 2, 10, -9};
int n = sizeof(arr)/sizeof(arr[0]);

oddEvenSort(arr, n);
printArray(arr, n);

return (0);
}
```

## Python3

```# Python Program to implement
# Odd-Even / Brick Sort

def oddEvenSort(arr, n):
# Initially array is unsorted
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 = ' ')

# Code Contribute by Mohit Gupta_OMG <(0_o)>
```

## Java

```// Java Program to implement
// Odd-Even / Brick Sort
import java.io.*;

class GFG
{
public static void oddEvenSort(int arr[], int n)
{
boolean isSorted = false; // Initially array is unsorted

while (!isSorted)
{
isSorted = true;
int temp =0;

// Perform Bubble sort on odd indexed element
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;
}
}

// Perform Bubble sort on even indexed element
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(" ");
}
}
// Code Contribute by Mohit Gupta_OMG <(0_o)>
```

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 ?

This article is contributed by Rachit Belwariar. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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