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

Algorithm in Scala

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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

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.

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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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