# Python program to get all unique combinations of two Lists

Last Updated : 14 Mar, 2023

The combination is a mathematical technique that calculates the number of possible arrangements in a collection of items or list. In combination order of selection doesn’t matter. The unique combination of two lists in Python can be formed by pairing each element of the first list with the elements of the second list.

Example:

List_1 = ["a","b"]
List_2 = [1,2]
Unique_combination = [[('a',1),('b',2)],[('a',2),('b',1)]]

Method 1: Using permutation() of itertools package and zip() function.

Approach :

• Import itertools package and initialize list_1 and list_2.
• Create an empty list of ‘unique_combinations’ to store the resulting combinations so obtained.
• Call itertools.permutations( ) which will return permutations of list_1 with length of list_2. Generally, the length of the shorter list is taken and if both lists are equal, use either.
• For loop is used and zip() function is called to pair each permutation and shorter list element into the combination.
• Then each combination is converted into a list and append to the combination list.

Below is the implementation.

## Python3

 # python program to demonstrate # unique combination of two lists # using zip() and permutation of itertools   # import itertools package import itertools from itertools import permutations   # initialize lists list_1 = ["a", "b", "c","d"] list_2 = [1,4,9]   # create empty list to store the # combinations unique_combinations = []   # Getting all permutations of list_1 # with length of list_2 permut = itertools.permutations(list_1, len(list_2))   # zip() is called to pair each permutation # and shorter list element into combination for comb in permut:     zipped = zip(comb, list_2)     unique_combinations.append(list(zipped))   # printing unique_combination list print(unique_combinations)

Output :

[[(‘a’, 1), (‘b’, 4), (‘c’, 9)], [(‘a’, 1), (‘b’, 4), (‘d’, 9)], [(‘a’, 1), (‘c’, 4), (‘b’, 9)], [(‘a’, 1), (‘c’, 4), (‘d’, 9)], [(‘a’, 1), (‘d’, 4), (‘b’, 9)], [(‘a’, 1), (‘d’, 4), (‘c’, 9)], [(‘b’, 1), (‘a’, 4), (‘c’, 9)], [(‘b’, 1), (‘a’, 4), (‘d’, 9)], [(‘b’, 1), (‘c’, 4), (‘a’, 9)], [(‘b’, 1), (‘c’, 4), (‘d’, 9)], [(‘b’, 1), (‘d’, 4), (‘a’, 9)], [(‘b’, 1), (‘d’, 4), (‘c’, 9)], [(‘c’, 1), (‘a’, 4), (‘b’, 9)], [(‘c’, 1), (‘a’, 4), (‘d’, 9)], [(‘c’, 1), (‘b’, 4), (‘a’, 9)], [(‘c’, 1), (‘b’, 4), (‘d’, 9)], [(‘c’, 1), (‘d’, 4), (‘a’, 9)], [(‘c’, 1), (‘d’, 4), (‘b’, 9)], [(‘d’, 1), (‘a’, 4), (‘b’, 9)], [(‘d’, 1), (‘a’, 4), (‘c’, 9)], [(‘d’, 1), (‘b’, 4), (‘a’, 9)], [(‘d’, 1), (‘b’, 4), (‘c’, 9)], [(‘d’, 1), (‘c’, 4), (‘a’, 9)], [(‘d’, 1), (‘c’, 4), (‘b’, 9)]]

Time complexity: O(n! * m), where n is the length of list_1 and m is the length of list_2.
Auxiliary Space: O(n! * m), as we are storing all the unique combinations in the “unique_combinations” list.

Method 2 : Using product() of itertools package and zip() function.

Approach :

• Import itertools package and initialize list_1 and list_2.
• Create an empty list of ‘unique_combinations’ to store the resulting combinations so obtained.
• product() is called to find all possible combinations of elements.
• And zip() is used to pair up all these combinations, converting each element into a list and append them to the desired combination list.

Below is the implementation.

## Python3

 # python program to demonstrate # unique combination of two lists # using zip() and product() of itertools   # import itertools package import itertools from itertools import product   # initialize lists list_1 = ["b","c","d"] list_2 = [1,4,9]   # create empty list to store the combinations unique_combinations = []   # Extract Combination Mapping in two lists # using zip() + product() unique_combinations = list(list(zip(list_1, element))                            for element in product(list_2, repeat = len(list_1)))   # printing unique_combination list print(unique_combinations)

Output :

[[(‘b’, 1), (‘c’, 1), (‘d’, 1)], [(‘b’, 1), (‘c’, 1), (‘d’, 4)], [(‘b’, 1), (‘c’, 1), (‘d’, 9)], [(‘b’, 1), (‘c’, 4), (‘d’, 1)], [(‘b’, 1), (‘c’, 4), (‘d’, 4)], [(‘b’, 1), (‘c’, 4), (‘d’, 9)], [(‘b’, 1), (‘c’, 9), (‘d’, 1)], [(‘b’, 1), (‘c’, 9), (‘d’, 4)], [(‘b’, 1), (‘c’, 9), (‘d’, 9)], [(‘b’, 4), (‘c’, 1), (‘d’, 1)], [(‘b’, 4), (‘c’, 1), (‘d’, 4)], [(‘b’, 4), (‘c’, 1), (‘d’, 9)], [(‘b’, 4), (‘c’, 4), (‘d’, 1)], [(‘b’, 4), (‘c’, 4), (‘d’, 4)], [(‘b’, 4), (‘c’, 4), (‘d’, 9)], [(‘b’, 4), (‘c’, 9), (‘d’, 1)], [(‘b’, 4), (‘c’, 9), (‘d’, 4)], [(‘b’, 4), (‘c’, 9), (‘d’, 9)], [(‘b’, 9), (‘c’, 1), (‘d’, 1)], [(‘b’, 9), (‘c’, 1), (‘d’, 4)], [(‘b’, 9), (‘c’, 1), (‘d’, 9)], [(‘b’, 9), (‘c’, 4), (‘d’, 1)], [(‘b’, 9), (‘c’, 4), (‘d’, 4)], [(‘b’, 9), (‘c’, 4), (‘d’, 9)], [(‘b’, 9), (‘c’, 9), (‘d’, 1)], [(‘b’, 9), (‘c’, 9), (‘d’, 4)], [(‘b’, 9), (‘c’, 9), (‘d’, 9)]]

Time complexity: O(n! * m), where n is the length of list_1 and m is the length of list_2.
Auxiliary Space: O(n! * m), as we are storing all the unique combinations in the “unique_combinations” list.

Method 3:use nested loops:

In this program generates all possible unique combinations of elements from two input lists. It uses nested loops to iterate over each element in both lists and adds them as a tuple to the unique_combinations list. The final output is a list of all possible combinations of elements from the two input list

## Python3

 list_1 = ["b", "c", "d"] list_2 = [1, 4, 9]   unique_combinations = []   for i in range(len(list_1)):     for j in range(len(list_2)):         unique_combinations.append((list_1[i], list_2[j]))   print(unique_combinations)

Output

[('b', 1), ('b', 4), ('b', 9), ('c', 1), ('c', 4), ('c', 9), ('d', 1), ('d', 4), ('d', 9)]

The time complexity of this method is O(n^2), where n is the length of the lists.

The space complexity is also O(n^2) because it creates a list of tuples that has the same number of elements as the product of the lengths of the input lists.

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