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Cocktail Sort

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Cocktail Sort is a variation ofBubble sort. The Bubble sort algorithm always traverses elements from left and moves the largest element to its correct position in first iteration and second largest in second iteration and so on. Cocktail Sort traverses through a given array in both directions alternatively.


Each iteration of the algorithm is broken up into 2 stages:

The first stage loops through the array from left to right, just like the Bubble Sort. During the loop, adjacent items are compared and if value on the left is greater than the value on the right, then values are swapped. At the end of first iteration, largest number will reside at the end of the array. The second stage loops through the array in opposite direction- starting from the item just before the most recently sorted item, and moving back to the start of the array. Here also, adjacent items are compared and are swapped if required.

Example :Let us consider an example array (5 1 4 2 8 0 2)

First Forward Pass:

( 5 1 4 2 8 0 2) → ( 1 5 4 2 8 0 2), Swap since 5 > 1

(1 5 4 2 8 0 2) → (1 4 5 2 8 0 2), Swap since 5 > 4

(1 4 5 2 8 0 2) → (1 4 2 5 8 0 2), Swap since 5 > 2

(1 4 2 5 8 0 2) → (1 4 2 5 8 0 2)

(1 4 2 5 8 0 2) → (1 4 2 5 0 8 2), Swap since 8 > 0

(1 4 2 5 0 8 2 ) → (1 4 2 5 0 2 8 ), Swap since 8 > 2

After first forward pass, greatest element of the array will be present at the last index of array.

First Backward Pass:

(1 4 2 5 0 2 8) → (1 4 2 5 0 2 8)

(1 4 2 5 0 2 8) → (1 4 2 0 5 2 8), Swap since 5 > 0

(1 4 2 0 5 2 8) → (1 4 0 2 5 2 8), Swap since 2 > 0

(1 4 0 2 5 2 8) → (1 0 4 2 5 2 8), Swap since 4 > 0

( 1 0 4 2 5 2 8) → ( 0 1 4 2 5 2 8), Swap since 1 > 0

After first backward pass, smallest element of the array will be present at the first index of the array.

Second Forward Pass:

(0 1 4 2 5 2 8) → (0 1 4 2 5 2 8)

(0 1 4 2 5 2 8) → (0 1 2 4 5 2 8), Swap since 4 > 2

(0 1 2 4 5 2 8) → (0 1 2 4 5 2 8)

(0 1 2 4 5 2 8) → (0 1 2 4 2 5 8), Swap since 5 > 2

Second Backward Pass:

(0 1 2 4 2 5 8) → (0 1 2 2 4 5 8), Swap since 4 > 2

Now, the array is already sorted, but our algorithm doesn’t know if it is completed. The algorithm needs to complete this whole pass without any swap to know it is sorted.

(0 1 2 2 4 5 8) → (0 1 2 2 4 5 8)

(0 1 2 2 4 5 8) → (0 1 2 2 4 5 8)

C++ // C++ implementation of Cocktail Sort
#include<bits/stdc++.h>
using namespace std;
// Sorts arrar a[0..n-1] using Cocktail sort
void CocktailSort(int a[], int n)
{
bool swapped = true;
int start = 0;
int end = n-1;
while (swapped)
{
// reset the swapped flag on entering
// the loop, because it might be true from
// a previous iteration.
swapped = false;
// loop from left to right same as
// the bubble sort
for (int i = start; i < end; ++i)
{
if (a[i] > a[i + 1])
{
swap(a[i], a[i+1]);
swapped = true;
}
}
// if nothing moved, then array is sorted.
if (!swapped)
break;
// otherwise, reset the swapped flag so that it
// can be used in the next stage
swapped = false;
// move the end point back by one, because
// item at the end is in its rightful spot
--end;
// from right to left, doing the
// same comparison as in the previous stage
for (int i = end - 1; i >= start; --i)
{
if (a[i] > a[i + 1])
{
swap(a[i], a[i+1]);
swapped = true;
}
}
// increase the starting point, because
// the last stage would have moved the next
// smallest number to its rightful spot.
++start;
}
}
/* Prints the array */
void printArray(int a[], int n)
{
for (int i=0; i<n; i++)
printf("%d ", a[i]);
printf("\n");
}
// Driver code
int main()
{
int arr[] = {5, 1, 4, 2, 8, 0, 2};
int n = sizeof(arr)/ sizeof(arr[0]);
CocktailSort(a,n);
printf("Sorted array :\n");
printArray(a,n);
return 0;
} Java // Java program for implementation of Cocktail Sort
public class CocktailSort
{
void cocktailSort(int a[])
{
boolean swapped = true;
int start = 0;
int end = a.length;
while (swapped==true)
{
// reset the swapped flag on entering the
// loop, because it might be true from a
// previous iteration.
swapped = false;
// loop from bottom to top same as
// the bubble sort
for (int i = start; i < end-1; ++i)
{
if (a[i] > a[i + 1])
{
int temp = a[i];
a[i] = a[i+1];
a[i+1] = temp;
swapped = true;
}
}
// if nothing moved, then array is sorted.
if (swapped==false)
break;
// otherwise, reset the swapped flag so that it
// can be used in the next stage
swapped = false;
// move the end point back by one, because
// item at the end is in its rightful spot
end = end-1;
// from top to bottom, doing the
// same comparison as in the previous stage
for (int i = end-1; i >=start; i--)
{
if (a[i] > a[i+1])
{
int temp = a[i];
a[i] = a[i+1];
a[i+1] = temp;
swapped = true;
}
}
// increase the starting point, because
// the last stage would have moved the next
// smallest number to its rightful spot.
start = start+1;
}
}
/* Prints the array */
void printArray(int a[])
{
int n = a.length;
for (int i=0; i<n; i++)
System.out.print(a[i] + " ");
System.out.println();
}
// Driver method
public static void main(String[] args)
{
CocktailSort ob = new CocktailSort();
int a[] = {5,1,4,2,8,0,2};
ob.cocktailSort(a);
System.out.println("Sorted array");
ob.printArray(a);
}
} python # Python program for implementation of Cocktail Sort
def cocktailSort(a):
n = len(a)
swapped = True
start = 0
end = n-1
while (swapped==True):
# reset the swapped flag on entering the loop,
# because it might be true from a previous
# iteration.
swapped = False
# loop from left to right same as the bubble
# sort
for i in range (start, end):
if (a[i] > a[i+1]) :
a[i], a[i+1]= a[i+1], a[i]
swapped=True
# if nothing moved, then array is sorted.
if (swapped==False):
break
# otherwise, reset the swapped flag so that it
# can be used in the next stage
swapped = False
# move the end point back by one, because
# item at the end is in its rightful spot
end = end-1
# from right to left, doing the same
# comparison as in the previous stage
for i in range(end-1, start-1,-1):
if (a[i] > a[i+1]):
a[i], a[i+1] = a[i+1], a[i]
swapped = True
# increase the starting point, because
# the last stage would have moved the next
# smallest number to its rightful spot.
start = start+1
# Driver code to test above
a = [5, 1, 4, 2, 8, 0, 2]
cocktailSort(a)
print("Sorted array is:")
for i in range(len(a)):
print ("%d" %a[i]),

Output:

Sorted array is:
0 1 2 2 4 5 8

Worst and Average Case Time Complexity:O(n*n).

Best Case Time Complexity:O(n). Best case occurs when array is already sorted.

Auxiliary Space:O(1)

Sorting In Place: Yes Stable: Yes

Comparison with Bubble Sort:

Time complexities are same, but Cocktail performs better than Bubble Sort. Typically cocktail sort is less than two times faster than bubble sort. Consider the example (2,3,4,5,1). Bubble sort of this example requires four traversals of array, while Cocktail sort requires only two traversals. (Source Wiki )

References: https://en.wikipedia.org/wiki/Cocktail_shaker_sort http://will.thimbleby.net/algorithms/doku.php?id=cocktail_sort http://www.programming-algorithms.net/article/40270/Shaker-sort

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

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.


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