Given a list, the task is to remove the duplicate elements. All the elements which are not in same order but made of same characters/numbers are considered as duplicates.
Examples:
Input : ['gfg', 'ggf', 'fgg', 'for', 'orf',
'ofr', 'rfo', 'rof', 'fro']
Output : ['for', 'fgg']
Input: ['110', '101', '001', '010', '100']
Output: ['001', '011']
Method #1 : Using set
Python3
# Python code to remove duplicate # unordered elements from a listfrom collections import Counter# List initialisationInput = ['1213','1231','1123','1132','2113', '2311','0007', '0016', '0025', '0034', '0043', '0052', '0061', '0070','0304', '0313', '0322','0098','9800', '0331', '0340', '0403', '0412', '0421', '0430', '0502','8900','8009' ,'0511', '0520', '0601', '0610', '0700', '1006', '1015']# Set initialisations = set()# Output list initialisationoutput =[]for i in Input: if tuple(Counter(sorted(i, key = int)).items()) in s: pass else: s.add(tuple(Counter(sorted(i, key = int)).items())) output.append(i)# Printing outputprint(output) |
Output:
['1213', '0007', '0016', '0025', '0034', '0313', '0322', '0098', '0412', '0511']
Time Complexity: O(n), where n is the length of the list
Auxiliary Space: O(n) additional space of size n is created where n is the number of elements in the list
Method #2 :
Python3
# Python code to remove duplicate# unordered elements from a list# List initialisationInput = ['gfg', 'ggf', 'fgg', 'for', 'orf', 'ofr', 'rfo', 'rof', 'fro']# Getting unique nosOutput = list({''.join(sorted(n)) for n in Input})# Printing Outputprint(Output) |
Output:
['for', 'fgg']
Method #3:Using Counter() method
Python3
# Python code to remove duplicate# unordered elements from a listfrom collections import Counter# List initialisationInput = ['gfg', 'ggf', 'fgg', 'for', 'orf', 'ofr', 'rfo', 'rof', 'fro']sortedList = list(''.join(sorted(n)) for n in Input)freq = Counter(sortedList)output = list(freq.keys())output.sort()# Printing Outputprint(output) |
Output
['fgg', 'for']
Time Complexity:O(NLogN)
Auxiliary Space: O(N)
