The job is simple. We need to take a list, with duplicate elements in it and generate another list that only contains the element without the duplicates in them.
Examples:
Input : [2, 4, 10, 20, 5, 2, 20, 4] Output : [2, 4, 10, 20, 5] Input : [28, 42, 28, 16, 90, 42, 42, 28] Output : [28, 42, 16, 90]
We can use not in on list to find out the duplicate items. We create a result list and insert only those that are not already not in.
Python3
# Python code to remove duplicate elementsdef Remove(duplicate): final_list = [] for num in duplicate: if num not in final_list: final_list.append(num) return final_list # Driver Codeduplicate = [2, 4, 10, 20, 5, 2, 20, 4]print(Remove(duplicate)) |
Output:
[2, 4, 10, 20, 5]
Easy Implementation:
A quick way to do the above using set data structure from the python standard library (Python 3.x implementation is given below)
Python3
duplicate = [2, 4, 10, 20, 5, 2, 20, 4]print(list(set(duplicate))) |
Output:
[2, 4, 10, 20, 5]
Method 2: Using Dictionary/hashmap
Approach:
- Create a dictionary and by default keep the count of every element to zero, using the default dict.
- If the count of elements is ZERO, increment the value in the dictionary and continue.
- If the count of element is greater than zero, Then remove the element from the given list using the remove() method.
You can read more about default dict here.
Python3
from collections import defaultdictdef default_val(): return 0# dict : maintain count of each element. with default value of key is 0mydict = defaultdict(default_val)# LISTl = [1, 2, 3, 2, 6, 3, 5, 3, 7, 8]for i in l: # if the element already present in the array, remove the element. if mydict[i] == 1: l.remove(i) # If the elements appears first time keep it count as 1 else: mydict[i] = 1# printing the final arrayprint(l) |
[1, 2, 6, 5, 3, 7, 8]
Method: Using dict.fromkeys()
Python3
input_list = [1, 2, 3, 2, 6, 3, 5, 3, 7, 8]mylist = list(dict.fromkeys(input_list))print(mylist) |
[1, 2, 3, 6, 5, 7, 8]
Time complexity: Average-case time complexity: O(n); n is the size of the array
Worst-case time complexity: O(
); n is the size of the array
Auxiliary Space: O(n); n is the size of the array
); n is the size of the arrayMethod: Using list comprehension:
Python
list1 = [1, 2, 3, 2, 6, 3, 5, 3, 7, 8]mylist = [ list1[i] for i in range(len(list1)) if list1.index(list1[i]) == i]print(mylist) |
Output:
[1, 2, 3, 6, 5, 7, 8]
Method: Using Counter() function
Python3
# Python code to remove duplicate elementsfrom collections import Counter# Driver Codeduplicate = [2, 4, 10, 20, 5, 2, 20, 4]unique = Counter(duplicate)print(list(unique.keys())) |
[2, 4, 10, 20, 5]
Time Complexity: O(N)
Auxiliary Space : O(N)
Method: Using operator.countOf() method
Python3
import operator as op# Python code to remove duplicate elementsdef Remove(duplicate): final_list = [] for num in duplicate: if op.countOf(final_list, num) == 0: final_list.append(num) return final_list# Driver Codeduplicate = [2, 4, 10, 20, 5, 2, 20, 4]print(Remove(duplicate)) |
[2, 4, 10, 20, 5]
Time Complexity: O(N)
Auxiliary Space : O(N)
