Python | Incremental slice partition in list
Last Updated :
26 Apr, 2023
Sometimes, while working with lists, we can come across a problem in which we need to slice a list incrementally, i.e., with each slice, the number of elements increases by 1. This has applications in competitive programming. Let’s discuss certain ways in which this task can be performed.
Method #1: Using loops
This is the brute force method by which this task can be performed. We just manually count and increase the counter with each iteration for slicing and dictionary key creation.
Python3
test_list = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
print("The original list is : " + str(test_list))
res = {}
N = 1
strt = 0
while strt < len(test_list):
res[N] = test_list[strt: strt + N]
strt += N
N += 1
print("The partitioned dictionary from list is : " + str(res))
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Output
The original list is : [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
The partitioned dictionary from list is : {1: [1], 2: [2, 3], 3: [4, 5, 6], 4: [7, 8, 9, 10]}
Time complexity: O(n^2)
Auxiliary space: O(n)
Method #2: Using enumerate() + slice() + next() + iter() + count()
The combination of the above functions can be used to perform this task. In this, next() is used to iterate the list converted to iterator by iter(). The slice() performs list slicing. The count() helps in managing the counter and enumerate keeps track of elements and indices in the list.
Python3
from itertools import islice, count
test_list = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
print("The original list is : " + str(test_list))
res = {key: val for key, val in enumerate(iter(lambda i=iter(test_list),
c=count(1): list(islice(i, next(c))), []), 1)}
print("The partitioned dictionary from list is : " + str(res))
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Output
The original list is : [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
The partitioned dictionary from list is : {1: [1], 2: [2, 3], 3: [4, 5, 6], 4: [7, 8, 9, 10]}
Time complexity: O(n)
Auxiliary space: O(n)
Using recursion:
Python3
def recursive_slice(lst, start, step, res):
if start >= len(lst):
return res
res[step] = lst[start: start + step]
return recursive_slice(lst, start + step, step + 1, res)
test_list = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
result = {}
print(recursive_slice(test_list, 0, 1, result))
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Output
{1: [1], 2: [2, 3], 3: [4, 5, 6], 4: [7, 8, 9, 10]}
Time complexity: O(n)
Auxiliary Space: O(n)
Method 4: Using itertools
Python3
import itertools
test_list = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
print("The original list is: " + str(test_list))
res = {}
start = 0
for i in itertools.count(1):
slice_len = min(i, len(test_list) - start)
if slice_len == 0:
break
res[i] = test_list[start: start + slice_len]
start += slice_len
print("The partitioned dictionary from list is: " + str(res))
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Output
The original list is: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
The partitioned dictionary from list is: {1: [1], 2: [2, 3], 3: [4, 5, 6], 4: [7, 8, 9, 10]}
Time complexity: O(n^2), where n is the length of the input list.
Auxiliary space: O(n), where n is the length of the input list.
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