Say, I got a list which length can be odd or even. Every iteration , I remove two items from the list. If there is one or no item at the end, I end the execution.

If I store (length list)/2 every loop, I will get e.g. [5,4,3...] for a list of length 10.

But what I want is [1,2,3,...] and I couldn't figure out a relationship between the list size and this sequence...

Is there a way to achieve this without having a variable storing the current count?

i.e. with `func :: [Int] -> [Int]`

only

Answer:

You can certainly do this without changing the type signature of `func :: [Int] -> [Int]`

: have `func`

call a different function, which takes an extra argument that is the counter you were talking about:

```
func :: [Int] -> [Int]
func = go 0
where go _ [] = []
go _ [x] = []
go n (a:b:xs) = n : a : b : go (succ n) xs
func [11,12..20]
[0,11,12,1,13,14,2,15,16,3,17,18,4,19,20]
```

If you are determined to do this without managing the counter variable yourself, you can do something more contorted, such as zipping your input list into 3-tuples `(a,b,n)`

, where `a`

and `b`

are pairs of items from your input list and `n`

comes from `[1,2..]`

.

```
pairs :: [a] -> [(a,a)]
pairs [] = []
pairs [_] = []
pairs (a:b:xs) = (a,b) : pairs xs
func' :: [Int] -> [Int]
func' xs = concat $ zipWith (\n (a,b) -> [n,a,b]) [1,2..] $ pairs xs
func' [11,12..20]
[1,11,12,2,13,14,3,15,16,4,17,18,5,19,20]
```

This is quite a bit less readable in my opinion, and I would suggest you just do the easy, obvious thing in my first snippet.

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