Given a Matrix, get all the rows with all the list elements.
Input : test_list = [[7, 6, 3, 2], [5, 6], [2, 1, 8], [6, 1, 2]], sub_list = [1, 2]
Output : [[2, 1, 8], [6, 1, 2]]
Explanation : Extracted lists have 1 and 2.Input : test_list = [[7, 6, 3, 2], [5, 6], [2, 1, 8], [6, 1, 2]], sub_list = [2, 6]
Output : [[7, 6, 3, 2], [6, 1, 2]]
Explanation : Extracted lists have 2 and 6.
Method #1: Using loop
In this, we iterate for each row from Matrix, and check for the presence of each list element, if the present row is returned as a result. If any element is not present, row is flagged off.
Python3
# Python3 code to demonstrate working of# Rows with all List elements# Using loop# initializing listtest_list = [[7, 6, 3, 2], [5, 6], [2, 1, 8], [6, 1, 2]]# printing original listprint("The original list is : " + str(test_list))# initializing listsub_list = [1, 2]res = []for row in test_list: flag = True # checking for all elements in list for ele in sub_list: if ele not in row: flag = False if flag: res.append(row)# printing resultprint("Rows with list elements : " + str(res)) |
Output:
The original list is : [[7, 6, 3, 2], [5, 6], [2, 1, 8], [6, 1, 2]] Rows with list elements : [[2, 1, 8], [6, 1, 2]]
Time Complexity: O(n*m)
Auxiliary Space: O(k)
Method #2 : Using all() + list comprehension
In this, all elements for presence as tested using all(), list comprehension is used as a one-liner to perform the task of iterating through rows.
Python3
# Python3 code to demonstrate working of# Rows with all List elements# Using all() + list comprehension# initializing listtest_list = [[7, 6, 3, 2], [5, 6], [2, 1, 8], [6, 1, 2]]# printing original listprint("The original list is : " + str(test_list))# initializing listsub_list = [1, 2]# testing elements presence using all()res = [row for row in test_list if all(ele in row for ele in sub_list)]# printing resultprint("Rows with list elements : " + str(res)) |
Output:
The original list is : [[7, 6, 3, 2], [5, 6], [2, 1, 8], [6, 1, 2]] Rows with list elements : [[2, 1, 8], [6, 1, 2]]
Time Complexity: O(n*m)
Auxiliary Space: O(k)
Method #3: Using nested loops
Use a nested loops to iterate through the rows in test_list and the elements in sub_list, checking if each element in sub_list is present in each row of test_list. If all elements in sub_list are present in a row, the row is added to the result list res.
Python3
# Python3 code to demonstrate working of# Rows with all List elements# Using nested loops# initializing listtest_list = [[7, 6, 3, 2], [5, 6], [2, 1, 8], [6, 1, 2]]# printing original listprint("The original list is : " + str(test_list))# initializing listsub_list = [1, 2]# initializing result listres = []# looping through the rows in the test_listfor row in test_list: # initializing a boolean variable to check if all elements in sub_list are in the row is_present = True # looping through the elements in sub_list for ele in sub_list: # if the element is not in the row, set is_present to False and break out of the loop if ele not in row: is_present = False break # if all elements in sub_list are present in the row, add the row to the result list if is_present: res.append(row)# printing resultprint("Rows with list elements : " + str(res)) |
The original list is : [[7, 6, 3, 2], [5, 6], [2, 1, 8], [6, 1, 2]] Rows with list elements : [[2, 1, 8], [6, 1, 2]]
Time complexity: O(n^2), where n is the number of rows in test_list.
Auxiliary space: O(n), where n is the number of rows in test_list.
Method #4: Using set intersection
Approach:
- Convert the sub_list to a set for faster intersection operation.
- Initialize an empty list, res, to store the rows that contain all the elements of sub_list.
- Iterate through each row of the test_list.
- Convert the current row to a set.
- Take the intersection of the current row set and the sub_list set using the & operator.
- If the length of the intersection is equal to the length of sub_list, then all the elements of sub_list are present in the current row.
- Append the current row to the res list if all the elements of sub_list are present in the current row.
- Return the res list as the result.
Python3
# Python3 code to demonstrate working of# Rows with all List elements# Using set intersection# initializing listtest_list = [[7, 6, 3, 2], [5, 6], [2, 1, 8], [6, 1, 2]]# printing original listprint("The original list is : " + str(test_list))# initializing listsub_list = [1, 2]sub_set = set(sub_list)# initializing result listres = []# iterating through each row of test_listfor row in test_list: # converting the row to a set row_set = set(row) # taking the intersection of row_set and sub_set intersection = row_set & sub_set # if all elements of sub_list are present in the row, append the row to res if len(intersection) == len(sub_set): res.append(row)# printing resultprint("Rows with list elements : " + str(res)) |
The original list is : [[7, 6, 3, 2], [5, 6], [2, 1, 8], [6, 1, 2]] Rows with list elements : [[2, 1, 8], [6, 1, 2]]
Time complexity: O(n*m), where n is the number of rows and m is the maximum length of a row in the test_list.
Auxiliary space: O(1), as we are not using any extra data structure except for the res list to store the result.
Method #5: Using set.issubset() and list comprehension
Approach:
- Initialize the original list “test_list”.
- Print the original list using the “print()” function.
- Initialize the sublist that we want to check for as “sub_list”.
- Initialize an empty list “res” to store the rows that have all the elements of the sublist.
- Use list comprehension to check if each row in the original list is a subset of the sublist. If a row is a subset of the sublist, append it to “res”.
- Print the list of rows that have all the elements of the sublist using the “print()” function.
Python3
# Python3 code to demonstrate working of# Rows with all List elements# Using set and all() function# initializing listtest_list = [[7, 6, 3, 2], [5, 6], [2, 1, 8], [6, 1, 2]]# printing original listprint("The original list is : " + str(test_list))# initializing sublist and setsub_list = [1, 2]sub_set = set(sub_list)# initializing result listres = []# iterating through each row of test_listfor row in test_list: # checking if sub_set is a subset of the set of elements in row if sub_set.issubset(set(row)): res.append(row)# printing resultprint("Rows with list elements : " + str(res)) |
The original list is : [[7, 6, 3, 2], [5, 6], [2, 1, 8], [6, 1, 2]] Rows with list elements : [[2, 1, 8], [6, 1, 2]]
Time complexity: O(n^2), where “n” is the number of elements in the original list.
Auxiliary space: O(m), where “m” is the number of rows in the original list that have all the elements of the sublist.
Method #6: Using map() and set() functions
Steps:
- Initialize the given list of lists test_list with the provided elements.
- Initialize the sub-list to find sub_list with the given values [1, 2].
- Use the built-in map() function to convert each inner list of test_list into a set.
- Use the built-in set() function to convert the sub_list into a set.
- Use the filter() function with a lambda function that checks if the set of sub_list is a subset of the set of the current row being filtered.
- Convert the filtered rows back into a list using the list() function.
- Print the final list of rows that contain all elements of the sub_list.
Python3
# Python3 code to demonstrate working of# Rows with all List elements# Using map() and set()# initializing listtest_list = [[7, 6, 3, 2], [5, 6], [2, 1, 8], [6, 1, 2]]# printing original listprint("The original list is : " + str(test_list))# initializing listsub_list = [1, 2]# finding rows that contain all elements of sub_list using map() and set()res = list(filter(lambda row: set(sub_list).issubset(set(row)), test_list))# printing resultprint("Rows with list elements : " + str(res)) |
The original list is : [[7, 6, 3, 2], [5, 6], [2, 1, 8], [6, 1, 2]] Rows with list elements : [[2, 1, 8], [6, 1, 2]]
Time complexity: O(NM), where N is the number of rows in the list and M is the length of the longest row.
Auxiliary space: O(NM), where N is the number of rows in the list and M is the length of the longest row.
