Sometimes, while working with list, we can have a problem in which we need to shift some sublist to the desired index in the same sublist. This problem can occur in day-day programming. Let’s discuss certain ways in which this task can be performed.
Method #1 : Using insert() + pop() + loop
The combination of above functions can be used to perform a particular task. The pop function can be used to remove the sublist and insert function inserts the sublist. This happens for each element in a single iteration in a loop.
Python3
# Python3 code to demonstrate working of# Shift sublist in list# Using insert() + pop() + loop# function to perform the taskdef shift_sublist(test_list, strt_idx, no_ele, shft_idx): for ele in range(no_ele): test_list.insert(shft_idx + 1, test_list.pop(strt_idx)) return test_list# initializing listtest_list = [4, 5, 6, 7, 3, 8, 10, 1, 12, 15, 16]# printing original listprint("The original list is : " + str(test_list))# Using insert() + pop() + loop# Shift sublist in listres = shift_sublist(test_list, 2, 3, 6)# Printing resultprint("The list after shift of sublist : " + str(res)) |
The original list is : [4, 5, 6, 7, 3, 8, 10, 1, 12, 15, 16] The list after shift of sublist : [4, 5, 8, 10, 1, 6, 7, 3, 12, 15, 16]
Time Complexity: O(n*n) where n is the number of elements in the list “test_list”.
Auxiliary Space: O(n), where n is the number of elements in the new res list
Method #2: Using list slicing
This task can also be done using a list slicing technique in which one can just add the different sections of list at the required positions.
Python3
# Python3 code to demonstrate working of# Shift sublist in list# Using list slicing# function to perform the taskdef shift_sublist(test_list, strt_idx, no_ele, shft_idx): return (test_list[:strt_idx] + test_list[strt_idx + no_ele : no_ele + shft_idx - 1] + test_list[strt_idx : strt_idx + no_ele] + test_list[shft_idx + no_ele -1:])# initializing listtest_list = [4, 5, 6, 7, 3, 8, 10, 1, 12, 15, 16]# printing original listprint("The original list is : " + str(test_list))# Using list slicing# Shift sublist in listres = shift_sublist(test_list, 2, 3, 6)# Printing resultprint("The list after shift of sublist : " + str(res)) |
The original list is : [4, 5, 6, 7, 3, 8, 10, 1, 12, 15, 16] The list after shift of sublist : [4, 5, 8, 10, 1, 6, 7, 3, 12, 15, 16]
Time Complexity: O(n) where n is the number of elements in the list “test_list”. Using list slicing performs n number of operations.
Auxiliary Space: O(n), where n is the length of the list.
Method #3: Using list comprehension and slicing
In this method, extract the sublist to be shifted using list slicing, create a new list by iterating through the original list and excluding the indices of the sublist, and insert the shifted sublist at the desired index using slicing.
Python3
def shift_sublist(test_list, strt_idx, no_ele, shft_idx): # Extract sublist to be shifted using list slicing and reverse it shifted_sublist = test_list[strt_idx:strt_idx+no_ele][::-1] # Create a new list by excluding indices of the sublist shifted_list = [test_list[i] for i in range(len(test_list)) if i not in range(strt_idx, strt_idx+no_ele)] # Insert the shifted sublist at the desired index using slicing shifted_list[shft_idx:no_ele+shft_idx] = shifted_sublist return shifted_list# initializing listtest_list = [4, 5, 6, 7, 3, 8, 10, 1, 12, 15, 16]# printing original listprint("The original list is:", test_list)# Using list comprehension and slicing# Shift sublist in listres = shift_sublist(test_list, 2, 3, 6)# Printing resultprint("The list after shift of sublist:", res) |
The original list is: [4, 5, 6, 7, 3, 8, 10, 1, 12, 15, 16] The list after shift of sublist: [4, 5, 8, 10, 1, 12, 3, 7, 6]
Time complexity: O(n), where n is the length of the original list. This is because we only need to iterate through the original list once to extract the shifted sublist and create a new list.
Auxiliary space: O(n), as we create a new list to hold the shifted sublist and a new list to hold the remaining elements of the original list.
Method #4: Using reduce():
- Initialize the result as the original list.
- For each shift in the list of shifts:
a. Call shift_sublist function on the result, passing in the current shift.
b. Update the result with the returned value from shift_sublist. - Return the final result.
Python3
from functools import reducedef shift_sublist(test_list, strt_idx, no_ele, shft_idx): return (test_list[:strt_idx] + test_list[strt_idx + no_ele : no_ele + shft_idx - 1] + test_list[strt_idx : strt_idx + no_ele] + test_list[shft_idx + no_ele - 1:])def shift_sublists(test_list, shifts): return reduce(lambda res, shift: shift_sublist(res, *shift), shifts, test_list)test_list = [4, 5, 6, 7, 3, 8, 10, 1, 12, 15, 16]shifts = [(2, 3, 6), (5, 2, 9), (1, 4, 8)]result = shift_sublists(test_list, shifts)print("The original list is : " + str(test_list))print("The list after shift of sublists : " + str(result))#This code is contributed by Jyothi pinjala. |
The original list is : [4, 5, 6, 7, 3, 8, 10, 1, 12, 15, 16] The list after shift of sublists : [4, 3, 12, 15, 6, 7, 16, 5, 8, 10, 1]
Time complexity: O(n), where n is the length of the input list. The shift_sublists() function applies shift_sublist() function to each tuple in shifts cumulatively using the reduce() function. Therefore, the time complexity of shift_sublists() function is O(mn), where m is the number of tuples in shifts.
Auxiliary space: O(n), where n is the length of the input list. The shift_sublists() function applies shift_sublist() function to each tuple in shifts cumulatively using the reduce() function. Therefore, the space complexity of shift_sublists() function is also O(n).
