Sometimes, we have a problem in which we need to deal with the grouping of elements. These groupings are comparatively easier while working with the databases, but using languages, this can be tricky. Let’s discuss certain ways to perform the grouping in Python by K.
Method #1 : Using loops This is the brute force method to perform this particular task. In this we use loops to have each number’s K’s place and add that number to designated list.
Python3
# Python3 code to demonstrate# K Division Grouping# using loops# initializing listtest_list = [3, 12, 13, 22, 25, 30]# printing original listprint("The original list : " + str(test_list))# initializing K K = 7# using loops# K Division Groupingres = []dec = -1for num in sorted(test_list): while num // K != dec: res.append([]) dec += 1 res[-1].append(num)# print resultprint("The list after grouping by K is : " + str(res)) |
The original list : [3, 12, 13, 22, 25, 30] The list after grouping by K is : [[3], [12, 13], [], [22, 25], [30]]
Time Complexity: O(n*n), where n is the number of elements in the list “test_list”.
Auxiliary Space: O(n*n), where n is the number of elements in the list “test_list”.
Method #2 : Using list comprehension + max() + min() Another method to perform this particular task in a shortened way is by using list comprehension. The min and max function specify the number of lists required as inner lists and rest of the task is performed inside the list comprehension.
Python3
test_list = [3, 12, 13, 22, 25, 30]K = 7res = []dec = -1for num in sorted(test_list): while num // K != dec: res.append([]) dec += 1 res[-1].append(num)print("The list after grouping by K is : " + str(res)) |
The original list : [3, 12, 13, 22, 25, 30] The list after grouping by K is : [[3], [12, 13], [], [22, 25], [30]]
Time Complexity: O(n*n), where n is the number of elements in the list “test_list”.
Auxiliary Space: O(n*n), where n is the number of elements in the list “test_list”.
Method #3: Using list comprehension + sorted() + range()
Step-by-step Algorithm :
- Initialize an empty list res of sublists, where the number of sublists is (max(test_list) // K) + 1.
- Iterate over the sorted elements of test_list:
a) Compute the floor division of the current element by K.
b) Append the current element to the sublist in res corresponding to the computed index. - Return the res list containing the grouped sublists.
Python3
test_list = [3, 12, 13, 22, 25, 30]K = 7print("The original list before grouping by K is : ", str(test_list))# Using list comprehensionres = [[] for _ in range((max(test_list) // K) + 1)]for num in sorted(test_list): res[num // K].append(num)# print resultprint("The list after grouping by K is : ", res) |
The original list before grouping by K is : [3, 12, 13, 22, 25, 30] The list after grouping by K is : [[3], [12, 13], [], [22, 25], [30]]
Complexity Analysis :
Time Complexity: O(n logn)
This is due to the sorting step. The iteration over the sorted elements takes linear time, which is dominated by the sorting time.
Space Complexity: O(m)
The space complexity of this algorithm is O(m), where m is the number of sublists in res. The number of sublists is (max(test_list) // K) + 1, which is proportional to the range of values in test_list. The space used by res is therefore proportional to the range of values in test_list. The space used by the other variables and temporary data structures is negligible.
