Python | Remove all duplicates and permutations in nested list

Given a nested list, the task is to remove all duplicates and permutations in that nested list.

Input:  [[-11, 0, 11], [-11, 11, 0], [-11, 0, 11], 
         [-11, 2, -11], [-11, 2, -11], [-11, -11, 2]]
Output:  {(-11, 0, 11), (-11, -11, 2)}

Input:  [[-1, 5, 3], [3, 5, 0], [-1, 5, 3], 
         [1, 3, 5], [-1, 3, 5], [5, -1, 3]]
Output:  {(1, 3, 5), (0, 3, 5), (-1, 3, 5)}

Code #1: Using Map 

Python3

# Python code to remove all duplicates
# and permutations in nested list

#Initialisation
listOfPermut = [[-11, 0, 11], [-11, 11, 0], [-11, 0, 11],
                [-11, 2, -11], [-11, -11, 2], [2, -11, -11]]

# Sorting tuple then removing
output = set(map(lambda x: tuple(sorted(x)),listOfPermut))

# printing output
print(output)

Output:

{(-11, 0, 11), (-11, -11, 2)}

Code #2:

Python3

# Python code to remove all duplicates
# and permutations in nested list

# Initialisation
input = [[-11, 0, 11], [-11, 11, 0], [-11, 2, -11],
         [-11, -11, 2], [2, -11, -11]]

# Sorting tuple then removing
output = set(tuple(sorted(x)) for x in input)

# printing output
print(output)

Output:

{(-11, 0, 11), (-11, -11, 2)}

Code #3: Using sort() and not in operator

Python3

# Python code to remove all duplicates
# and permutations in nested list

# Initialisation
input = [[-11, 0, 11], [-11, 11, 0], [-11, 2, -11],
         [-11, -11, 2], [2, -11, -11]]

# Sorting tuple then removing
res = []
for i in input:
    i.sort()
    res.append(i)
output = []
for i in res:
    if i not in output:
        output.append(i)
output = list(map(tuple, output))

print(tuple(output))

Output

((-11, 0, 11), (-11, -11, 2))

Code #4: Using operator.countOf() method

Python3

# Python code to remove all duplicates
# and permutations in nested list
import operator as op
# Initialisation
input = [[-11, 0, 11], [-11, 11, 0], [-11, 2, -11],
         [-11, -11, 2], [2, -11, -11]]

# Sorting tuple then removing
res = []
for i in input:
    i.sort()
    res.append(i)
output = []
for i in res:
    if op.countOf(output, i) == 0:
        output.append(i)
output = list(map(tuple, output))

print(tuple(output))

Output

((-11, 0, 11), (-11, -11, 2))

Time Complexity: O(N*N)

Auxiliary Space: O(N*N)