# Solved: remove nth element from list

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In the world of functional programming and Haskell, dealing with lists is often at the core of many computational problems. One such problem involves removing an nth element from a list. This task comes up quite frequently, due to the inherent use of lists in expressing various data structures and computations.

Functional programming is a style of programming that treats computation as the evaluation of mathematical functions and avoids changing state and mutable data. In the realm of functional programming, Haskell holds a significant position and is immensely popular due to its strong static typing, high-level abstract nature, and pure functional approach, which makes it easier to ensure the accuracy of code and minimize errors.

One of the features that make Haskell distinctive and efficient is its inherent support for lists. Virtually any data can be expressed as a list, and many computations can also be represented using lists. Lists are homogenous data structures in Haskell which represents a collection of elements of the same type.

## Removing nth Element from a list in Haskell

In functional programming, we don’t alter or mutate the original data structure, but we usually create a new one with the desired changes. The same logic applies when we want to remove an nth element from a list in Haskell.

```removeNthElement :: Int -> [a] -> [a]
removeNthElement _ [] = []
removeNthElement n xs = take (n-1) xs ++ drop n xs
```

The function `removeNthElement` takes an integer ‘n’ and a list ‘xs’ as its input. The function then uses the ‘take’ function, which extracts the first ‘n-1’ elements, and the ‘drop’ function, which skips the first ‘n’ elements of the list. The operator ‘++’ then combines these two lists to form a new list from which the nth element is removed.