# Python – Cross List Sync on duplicate elements removal

• Last Updated : 21 Jan, 2023

Given two lists, perform syncing of all the elements of list with other list when removal of duplicate elements in first list.

Input : test_list1 = [1, 1, 2, 2, 3, 3], test_list2 = [8, 3, 7, 5, 4, 1]
Output : [1, 2, 3], [8, 7, 4]
Explanation : After removal of duplicates (1, 2, 3), corresponding elements are removed from 2nd list, and hence (8, 7, 4) are mapped.
Input : test_list1 = [1, 2, 2, 2, 2], test_list2 = [8, 3, 7, 5, 4, 1]
Output : [1, 2], [8, 3]
Explanation : Elements removed are (2, 2, 2), [1, 2] are mapped to [8, 3] as expected.

Method #1 : Using loop + dict()
The above functions provide brute force way to solve this problem. In this, we memorize the elements with its counterpart using dictionary and loop conditionals.

## Python3

 # Python3 code to demonstrate working of# Cross List Sync on duplicate elements removal# Using loop + dict() # initializing liststest_list1 = [2, 2, 3, 4, 4, 4, 5, 5, 6, 6]test_list2 = [8, 3, 7, 5, 4, 1, 0, 9, 4, 2] # printing original listsprint("The original list 1 : " + str(test_list1))print("The original list 2 : " + str(test_list2)) # Cross List Sync on duplicate elements removaltemp = dict()a = []for idx in range(len(test_list1)):    if test_list1[idx] not in a:        a.append(test_list1[idx])                 # performing memoize using dictionary        temp[test_list1[idx]] = test_list2[idx] res2 = list(temp.values())res1 = a # printing resultprint("List 1 : " + str(res1))print("Sync List : " + str(res2))

Output :

The original list 1 : [2, 2, 3, 4, 4, 4, 5, 5, 6, 6]
The original list 2 : [8, 3, 7, 5, 4, 1, 0, 9, 4, 2]
List 1 : [2, 3, 4, 5, 6]
Sync List : [8, 7, 5, 0, 4]

Method #2 : Using loop + zip()
This is yet another way in which this task can be performed. In this, we sync both the lists using zip(). The similar memorize logic is implemented as above task and corresponding valid other list element is appended to result.

## Python3

 # Python3 code to demonstrate working of# Cross List Sync on duplicate elements removal# Using loop + zip() # initializing liststest_list1 = [2, 2, 3, 4, 4, 4, 5, 5, 6, 6]test_list2 = [8, 3, 7, 5, 4, 1, 0, 9, 4, 2] # printing original listsprint("The original list 1 : " + str(test_list1))print("The original list 2 : " + str(test_list2)) # Cross List Sync on duplicate elements removalres1 = []res2 = [] # both lists are combined index wise using zip()for a, b in zip(test_list1, test_list2):    if a not in res1:        res1.append(a)        res2.append(b)         # printing resultprint("List 1 : " + str(res1))print("Sync List : " + str(res2))

Output :

The original list 1 : [2, 2, 3, 4, 4, 4, 5, 5, 6, 6]
The original list 2 : [8, 3, 7, 5, 4, 1, 0, 9, 4, 2]
List 1 : [2, 3, 4, 5, 6]
Sync List : [8, 7, 5, 0, 4]

Method #3:Using operator.countOf() method

## Python3

 # Python3 code to demonstrate working of# Cross List Sync on duplicate elements removalimport operator as op# initializing liststest_list1 = [2, 2, 3, 4, 4, 4, 5, 5, 6, 6]test_list2 = [8, 3, 7, 5, 4, 1, 0, 9, 4, 2] # printing original listsprint("The original list 1 : " + str(test_list1))print("The original list 2 : " + str(test_list2)) # Cross List Sync on duplicate elements removaltemp = dict()a = []for idx in range(len(test_list1)):    if op.countOf(a, test_list1[idx]) == 0:        a.append(test_list1[idx])         # performing memoize using dictionary        temp[test_list1[idx]] = test_list2[idx] res2 = list(temp.values())res1 = a # printing resultprint("List 1 : " + str(res1))print("Sync List : " + str(res2))

Output

The original list 1 : [2, 2, 3, 4, 4, 4, 5, 5, 6, 6]
The original list 2 : [8, 3, 7, 5, 4, 1, 0, 9, 4, 2]
List 1 : [2, 3, 4, 5, 6]
Sync List : [8, 7, 5, 0, 4]

Time Complexity:O(N)

Auxiliary Space: O(N)

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