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:
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 = okergsf With Order = geksfor
Time complexity: O(n)
Auxiliary Space: O(n)
Method 2:
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: soerkgf 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 :
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 :
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)
Method 5: Using operator.countOf() Method
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: rfogesk With Order: geksfor
Time Complexity: O(N)
Auxiliary Space : O(N)
