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Remove all duplicates from a given string in Python

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  • Difficulty Level : Easy
  • Last Updated : 21 Jul, 2022
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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 : efgkors

This problem has existing solution please refer 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 value and returns list of keys, fromkeys(seq[, value]) is the syntax for fromkeys() method. Parameters :

  • seq : This is the list of values which would be used for dictionary keys preparation.
  • value : This is optional, if provided then 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)

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. 


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