Given a list of lists, write a Python program to remove sublists from the given list of lists that are present in another sublist.
Examples:
Input : [['a', 'b', 'c'], ['a', 'c'], ['a', 'b', 'c'], ['d']] Output : [['a', 'b', 'c'], ['d']] Input : [[1], [1, 2], [1, 2, 3], [0], [0, 1]] Output : [[1, 2, 3], [0, 1]]
Approach #1: Using Python Set (If order of list doesn’t matter) This approach makes use of Python sets. Create two empty lists ‘curr_res’ to store current sublist and ‘result’ to store the finalized sublists. Convert the sub-lists in the given list of lists to sets and sort them by length in reverse order, so that you can iterate through them and add each set to the curr_res only if it is not a subset of any of the existing sets in the curr_res. The only drawback of this approach is that it may produce the result in an unordered way(Since sets are unordered).
Python3
# Python3 program to remove sublists from# list of lists that are in another sublistdef removeSublist(lst): curr_res = [] result = [] for ele in sorted(map(set, lst), key = len, reverse = True): if not any(ele <= req for req in curr_res): curr_res.append(ele) result.append(list(ele)) return result # Driver codelst = [['a', 'b', 'c'], ['a', 'b'], ['a', 'b', 'c'], ['d']]print(removeSublist(lst)) |
[['c', 'b', 'a'], ['d']]
Time Complexity: O(n)
Auxiliary Space: O(n)
Approach #2: Using Python Dictionary (If order of list matters) Dict may not always produce ordered output, therefore you can use OrderedDict from collections module.
Python3
# Python3 program to remove sublists from# list of lists that are in another sublistfrom collections import OrderedDictdef removeSublist(lst): curr_result = [] result = [] for ele in sorted(map(OrderedDict.fromkeys, lst), key = len, reverse = True): if not any(ele.keys() <= req.keys() for req in curr_result): curr_result.append(ele) result.append(list(ele)) return result # Driver codelst = [['a', 'b', 'c'], ['a', 'b'], ['a', 'b', 'c'], ['d']]print(removeSublist(lst)) |
[['a', 'b', 'c'], ['d']]
Time complexity: O(n^2), where n is the length of the input list of sublists.
Auxiliary space: O(n^2), because it creates a list of OrderedDicts called curr_result that can hold up to n^2 key-value pairs (if all sublists have distinct elements).
Approach #3: Using issubset()+list comprehension+enumerate()
This code defines a function remove_duplicate_sublists() that removes duplicate sublists and sublists that are present in another sublist from a list of lists. It first uses a list comprehension to create a new list with unique sublists by iterating over the sublists in the original list and checking if each sublist is not in the preceding sublists. It then uses a second list comprehension to filter out the sublists that are present in another sublist by using the issubset() method of the set() function. The resulting list is returned as the result.
Python3
def remove_duplicate_sublists(lists): # Use list comprehension to create a new list with unique sublists unique_lists = [sublist for i, sublist in enumerate(lists) if sublist not in lists[:i]] result = [sublist for sublist in unique_lists if not any(set(sublist).issubset(other) for other in unique_lists if other != sublist)] return resultlists = [['a', 'b', 'c'], ['a', 'c'], ['a', 'b', 'c'], ['d'], ['d']]print(remove_duplicate_sublists(lists)) # Output: [['a', 'b', 'c'], ['d']]lists = [[1], [1, 2], [1, 2, 3], [0], [0, 1], [1, 2]]print(remove_duplicate_sublists(lists)) # Output: [[1, 2, 3], [0, 1]]#This code is contributed by Edula Vinay Kumar Reddy |
[['a', 'b', 'c'], ['d']] [[1, 2, 3], [0, 1]]
Time complexity: O(n^2), because it involves nested loops over the sublists in the list of lists.
Auxiliary space: O(n), because it involves storing a list of n sublists in memory.
