README.md

Array.diff

Your goal in this kata is to implement a difference function, which subtracts one list from another and returns the result.

It should remove all values from list a, which are present in list b keeping their order.

array_diff({1, 2}, 2, {1}, 1, *z) == {2} (z == 1)

arrayDiff([1,2],[1]) == [2]

array_diff([1,2],[1]) == [2]

array_diff([1,2],[1]) == [2]

array_diff([1,2],[1]) == [2]

arrayDiff([1,2],[1]) == [2]

difference [1,2] [1] == [2]

Kata.ArrayDiff(new int[] {1, 2}, new int[] {1}) => new int[] {2}

arrayDiff [|1|] [|1; 2|] = [|2|]

array_diff(vec![1,2], vec![1]) == vec![2]

(= (array-diff [1 2] [1]) [2])

Kata.arrayDiff([1,2],[1]) == [2]

Kata.arrayDiff(new int[] {1, 2}, new int[] {1}) => new int[] {2}

arraydiff([1,2],[1]) == [2]

arrayDiff(@[1,2],@[1]) == @[2]

arrayDiff([1,2],[1]) == [2]

array_diff(c(1, 2), 1) == 2

array_diff([1, 2], [1], [2]). % Result = [2]

arrayDiff(Seq(1, 2), Seq(1)) == Seq(2)

      ArrayDiff([1, 2], [1]) = [2]

If a value is present in b, all of its occurrences must be removed from the other:

array_diff({1, 2, 2, 2, 3}, 5, {2}, 1, *z) == {1, 3} (z == 2)

arrayDiff([1,2,2,2,3],[2]) == [1,3]

array_diff([1,2],[1]) == [2]

array_diff([1,2,2,2,3],[2]) == [1,3]

arrayDiff([1,2,2,2,3],[2]) == [1,3]

difference [1,2,2,2,3] [2] == [1,3]

Kata.ArrayDiff(new int[] {1, 2, 2, 2, 3}, new int[] {2}) => new int[] {1, 3}

arrayDiff [|2|] [|1; 2; 2; 2; 3|] = [|1; 3|]

array_diff(vec![1,2,2,2,3], vec![2]) == vec![1,3]

(= (array-diff [1,2,2,2,3] [2]) [1,3])

Kata.arrayDiff([1,2,2,2,3],[2]) == [1,3]

Kata.arrayDiff(new int[] {1, 2, 2, 2, 3}, new int[] {2}) => new int[] {1, 3}

arraydiff([1,2,2,2,3],[2]) == [1,3]

arrayDiff(@[1,2,2,2,3],@[2]) == @[1,3]

arrayDiff([1,2,2,2,3],[2]) == [1,3]

array_diff(c(1, 2, 2, 2, 3), 2) == c(1, 3)

array_diff([1, 2, 2, 2, 3], [2], [1, 3]). % Result = [1, 3]

arrayDiff(Seq(1, 2, 2, 2, 2, 2, 3), Seq(2)) == Seq(1, 3)

      ArrayDiff([1,2,2,2,3],[2]) = [1,3]

NOTE: In C, assign return array length to pointer *z