# Problem with rearanging list with paired letters next to each others

```On 2020-11-11 01:26, Bischoop wrote:
>
> Can anybody help?Here's the code that gives me a hedeache for second day https://bpa.st/XLOA
> If I have letter_list = ["a","a",,"b","b","c","c"] it's working
> correctly but if I have three or more of any letters for
> example:letter_list = ["a","a","a","b","b","c","c"], I'm ending up with
> some pairs somewhere.
> I took another way approach today: https://bpa.st/E7HQ, was thinking
> about iterating and checking if neighbour characters won't make a pair
> but I've end up with error:
> if x != letter_list[i+1] and letter_list[i-1]:
> IndexError: list index out of range
> andin addition: got 4 "a" (had 3only) and missing 1 "b" :-/
>
Points to note in your first code:

1. Modifying a list while iterating over it is a bad idea.

2. You're modifying the list that you're passing in and also returning
it. That's a bad idea. Either modify it in place or modify and return a
copy.

3. The line:

m = (letter_list.index(x))

returns the position of the first occurrence of x.

Points to note in your second code:

1. (See above)

2. (See above)

3. (See above)

4. 'i' goes from 0 to len(letter_list)-1, so i+1 goes from 1 to
len(letter_list), but the maximum index permitted by letter_list is
len(letter_list)-1, hence letter_list[i+1] will raise IndexError on the
last iteration.

5. 'i' goes from 0 to len(letter_list)-1, so i-1 goes from -1 to
len(letter_list)-2. letter_list[-1] returns the last (final) letter in
the list, and it's treated as a true.

6. This:

x != letter_list[i+1] and letter_list[i-1]

is checking whether:

x != letter_list[i+1]

is true and also whether:

letter_list[i-1]

is true.

```