Remove all duplicates from a given string in Python

We are given a string and we need to remove all duplicates from it? What will be the output if the order of character matters? Examples:

Input : geeksforgeeks 
Output : geksfor

This problem has an existing solution please refer to Remove all duplicates from a given string. 

Method 1:

Python3

from collections import OrderedDict

# Function to remove all duplicates from string
# and order does not matter
def removeDupWithoutOrder(str):

    # set() --> A Set is an unordered collection
    #         data type that is iterable, mutable,
    #         and has no duplicate elements.
    # "".join() --> It joins two adjacent elements in
    #             iterable with any symbol defined in
    #             "" ( double quotes ) and returns a
    #             single string
    return "".join(set(str))

# Function to remove all duplicates from string
# and keep the order of characters same
def removeDupWithOrder(str):
    return "".join(OrderedDict.fromkeys(str))

# Driver program
if __name__ == "__main__":
    str = "geeksforgeeks"
    print ("Without Order = ",removeDupWithoutOrder(str))
    print ("With Order = ",removeDupWithOrder(str))

Output

Without Order =  foskerg
With Order =  geksfor

Time complexity: O(n)
Auxiliary Space: O(n)

Method 2:

Python3

def removeDuplicate(str):
    s=set(str)
    s="".join(s)
    print("Without Order:",s)
    t=""
    for i in str:
        if(i in t):
            pass
        else:
            t=t+i
        print("With Order:",t)

str="geeksforgeeks"
removeDuplicate(str)

Output

Without Order: kogerfs
With Order: g
With Order: ge
With Order: ge
With Order: gek
With Order: geks
With Order: geksf
With Order: geksfo
With Order: geksfor
With Order: geksfor
With Order: geksfor
With Order: geksfor
With Order: geksfor
With Order: geksfor

Time complexity: O(n)
Auxiliary Space: O(n)

What do OrderedDict and fromkeys() do?

An OrderedDict is a dictionary that remembers the order of the keys that were inserted first. If a new entry overwrites an existing entry, the original insertion position is left unchanged.

For example, see below code snippet : 

Python3

from collections import OrderedDict

ordinary_dictionary = {}
ordinary_dictionary['a'] = 1
ordinary_dictionary['b'] = 2
ordinary_dictionary['c'] = 3
ordinary_dictionary['d'] = 4
ordinary_dictionary['e'] = 5

# Output = {'a': 1, 'c': 3, 'b': 2, 'e': 5, 'd': 4}
print (ordinary_dictionary)

ordered_dictionary = OrderedDict()
ordered_dictionary['a'] = 1
ordered_dictionary['b'] = 2
ordered_dictionary['c'] = 3
ordered_dictionary['d'] = 4
ordered_dictionary['e'] = 5

# Output = {'a':1,'b':2,'c':3,'d':4,'e':5}
print (ordered_dictionary)

Output

{'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5}
OrderedDict([('a', 1), ('b', 2), ('c', 3), ('d', 4), ('e', 5)])

Time complexity: O(n)
Auxiliary Space: O(1)

fromkeys() creates a new dictionary with keys from seq and values set to a value and returns a list of keys, fromkeys(seq[, value]) is the syntax for fromkeys() method. Parameters :

• seq : This is the list of values that would be used for dictionary keys preparation.
• value : This is optional, if provided then the value would be set to this value.

For example, see below code snippet : 

Python3

from collections import OrderedDict
seq = ('name', 'age', 'gender')
dict = OrderedDict.fromkeys(seq)

# Output = {'age': None, 'name': None, 'gender': None}
print (str(dict))
dict = OrderedDict.fromkeys(seq, 10)

# Output = {'age': 10, 'name': 10, 'gender': 10}
print (str(dict))

Output

OrderedDict([('name', None), ('age', None), ('gender', None)])
OrderedDict([('name', 10), ('age', 10), ('gender', 10)])

Time complexity: O(n)
Auxiliary Space: O(1)

If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks. 

Method 5:  using operator.countOf() method

Python3

import operator as op


def removeDuplicate(str):
    s = set(str)
    s = "".join(s)
    print("Without Order:", s)
    t = ""
    for i in str:
        if op.countOf(t, i) > 0:
            pass
        else:
            t = t+i
        print("With Order:", t)


str = "geeksforgeeks"
removeDuplicate(str)

Output

Without Order: goksefr
With Order: g
With Order: ge
With Order: ge
With Order: gek
With Order: geks
With Order: geksf
With Order: geksfo
With Order: geksfor
With Order: geksfor
With Order: geksfor
With Order: geksfor
With Order: geksfor
With Order: geksfor

Time Complexity: O(N)

Auxiliary Space : O(N)