Hi, could you pease tell me the big O notation for this function? Thanks! # takes in an array as an argumentdefcocktail_sort(my_data):num_ops = 01n = Len(my_data)num_ops += 4for i in range(n):num_ops += 6swapped = Falsefor j in range(0, n - i - 1):num_ops += 6if my_data[j] > my_data[j + 1]:my_data[j], my_data[j + 1] = my_data[j + 1], my_data[j]swapped = Truenum_ops += 7for k in range (n i - 1, 0, -1):num_ops += 6if my_data[k] < my_data[k 1]:my_data[k], my_data[kswapped = Truenum_ops += 7num_ops += 3if (not swapped):break# print (num_ops)return my_data-1] = my_data[k - 1], my_data[k]

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the big O notation for this function

EBK JAVA PROGRAMMING

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ISBN:9781337671385

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Publisher:FARRELL

Chapter8: Arrays

Section: Chapter Questions

Problem 9PE

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Question

Hi, could you pease tell me the big O notation for this function? Thanks!

# takes in an array as an argument
def
cocktail_sort(my_data):
num_ops = 0
1
n = Len(my_data)
num_ops += 4
for i in range(n):
num_ops += 6
swapped = False
for j in range(0, n - i - 1):
num_ops += 6
if my_data[j] > my_data[j + 1]:
my_data[j], my_data[j + 1] = my_data[j + 1], my_data[j]
swapped = True
num_ops += 7
for k in range (n i - 1, 0, -1):
num_ops += 6
if my_data[k] < my_data[k 1]:
my_data[k], my_data[k
swapped = True
num_ops += 7
num_ops += 3
if (not swapped):
break
# print (num_ops)
return my_data
-
1] = my_data[k - 1], my_data[k]

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