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# 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 matterdef 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 samedef removeDupWithOrder(str):    return "".join(OrderedDict.fromkeys(str)) # Driver programif __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'] = 1ordinary_dictionary['b'] = 2ordinary_dictionary['c'] = 3ordinary_dictionary['d'] = 4ordinary_dictionary['e'] = 5 # Output = {'a': 1, 'c': 3, 'b': 2, 'e': 5, 'd': 4}print (ordinary_dictionary)     ordered_dictionary = OrderedDict()ordered_dictionary['a'] = 1ordered_dictionary['b'] = 2ordered_dictionary['c'] = 3ordered_dictionary['d'] = 4ordered_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 OrderedDictseq = ('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)

This article is contributed by Shashank Mishra (Gullu). 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)