Move items of a list in a cyclic way

Question

I have a sorted list. I want to move the values d steps forwards. And I need O(n) time complexity.

For example if d = 2, the value 1, which was in index = 0, now is in index = 2:

lst = [1,2,3,4,5]
d = 2

Expected output: [4,5,1,2,3]

The code:

def cyclic_lst(lst,d):
    mylist = []
    counter = 0
    while(counter <= d):
         mylist = lst[1:] + lst[0]
         counter += 1
    return mylist

I'm getting this error:

TypeError: can only concatenate list (not "int") to list

I can't use for loop with the while loop because of the complexity demand, how can I do this?

Your error can be fixed by changing the 6th line to `mylist = lst[1:] + [lst[0]]`, but this would be a band-aid on a gaping wound. — user19642323, Jul 30 '22 at 12:52
Because your program would still not do what you want it to. And I'm reluctant to write a full solution for you because this sounds a lot like a homework question... — user19642323, Jul 30 '22 at 12:56
@user19642323 it's not, it's from an exam from previous years. I'll be glad if you could give me a hint or something.. — Programming Noob, Jul 30 '22 at 12:57

Colim · Answer 1 · 2022-07-30T13:27:01.963

2

def cyclic_lst(lst,d):
    return lst[d:] + lst[:d]

But there is no need for cyclic_lst() at all: lst[idx-d-1] gives the same result, witouth calling any function or editing lists.

print(cyclic_lst(lst,d))
[3, 4, 5, 1, 2]

print(lst[0-d-1])
3
print(lst[1-d-1])
4
print(lst[2-d-1])
5
print(lst[3-d-1])
1
print(lst[4-d-1])
2

edited Jul 30 '22 at 13:27

answered Jul 30 '22 at 12:59

Colim

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score 1 · Answer 2 · answered Jul 30 '22 at 12:58

1

Here is a way to get the desired result:

lst = [1,2,3,4,5]
d = 2
new_lst = [lst[(i - d) % len(lst)] for i in range(len(lst))]

Output:

[4, 5, 1, 2, 3]

answered Jul 30 '22 at 12:58

constantstranger

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score 0 · Answer 3 · answered Jul 30 '22 at 13:04

Here's another way to do it, using the stylistic conventions of the original post.

def cyclic_lst(lst,d):
    mylist = []
    counter = 0
    while(counter <= d):
         mylist = lst[1:]
         mylist.append(lst[0])
         counter += 1
         lst = mylist
    return mylist

I modified your original algorithm slightly, and substituted the variable "mylst" back into "lst" in the while loop.

But this is really quite inefficient, and can't possibly be the best way of solving this problem.

I am not sure if I have misinterpreted the question or I am missing something here, but the variable new_lst can simply be obtained by slicing "lst" and joining the strings together. The simpler implementation below is O(n)

lst[d+1:] + lst[0:d+1]

MoRe · Answer 4 · 2022-07-31T14:10:46.730

0

lst[-d%len(lst):] + lst[:-d%len(lst)]

edited Jul 31 '22 at 14:10

answered Jul 30 '22 at 13:11

MoRe

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Move items of a list in a cyclic way

4 Answers4