# Scala algorithm: Is an Array a permutation?

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## Algorithm goal

A permutation of an array $$X$$ is another array of the same size, which has the same elements. $$[3,2,1],$$ is a permutation of $$[1,2,3]$$, but $$[2,1]$$ is not because $$3$$is missing.

Problem: For an array size $$N$$, check whether it is a permutation of $$[1, 2, ..., N]$$.

For example, for $$[4,3,2,1]$$, $$N = 4$$,and it is a permutation of sequence $$[1,2,3,4=N]$$.

However, for array $$[4,3,2]$$, $$N = 3$$, and $$[4,3,2]$$ is not a permutation of $$[1,2,3=N]$$.

Check whether an input is a permutation of $$(1..N)$$ where $$N$$ is the length of the input.

This problem is related to the Codility problem 'PermCheck'.

## Test cases in Scala

assert(isAPermutation(Array(1, 2)))
assert(!isAPermutation(Array(2)))
assert(isAPermutation(Array(2, 3, 4, 1)))
assert(!isAPermutation(Array(4, 1, 2)))
assert(!isAPermutation(Array(1, 2, 3, 3)))


## Algorithm in Scala

4 lines of Scala (version 2.13), showing how concise Scala can be!

## Explanation

The solution is very small yet intricate because Scala provides very terse syntax.

Firstly, the nature of the algorithm is: the length of the array corresponds to number $$N$$. Then we need to check that each item in $$(1..N)$$ is in the array. For this we use Scala's range syntax which allows us to produce a description (not a collection) of the range (which can then be acted on as a collection if need be). (this is © from www.scala-algorithms.com)

To check each item directly in the array with .contains would be very expensive as we'd have to go through the array mutliple times, leading to $$O(n^2)$$ complexity.

## Scala concepts & Hints

1. ### Range

The (1 to n) syntax produces a "Range" which is a representation of a sequence of numbers.

assert((1 to 5).toString == "Range 1 to 5")

assert((1 to 5).reverse.toString() == "Range 5 to 1 by -1")

assert((1 to 5).toList == List(1, 2, 3, 4, 5))


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