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Array.diff

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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.

array_diff([1,2],[1]) == [2]
array_diff([1,2],[1]) == [2]
array_diff([1,2],[1]) == [2]
array_diff([1,2],[1]) == [2]
difference [1,2] [1] == [2]
Kata.ArrayDiff(new int[] {1, 2}, new int[] {1}) => new int[] {2}
array_diff(vec![1,2], vec![1]) == vec![2]
(= (array-diff [1 2] [1]) [2])
Kata.arrayDiff([1,2],[1]) == [2]
arraydiff([1,2],[1]) == [2]
arrayDiff(@[1,2],@[1]) == @[2]
arrayDiff([1,2],[1]) == [2]
array_diff(c(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],[2]) == [1,3]
array_diff([1,2],[1]) == [2]
array_diff([1,2,2,2,3],[2]) == [1,3]
array_diff([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}
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]
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)

Timeline

  • Created: 2013-09-22
  • Published: 2013-09-22
  • Approved: null
  • Completed: 2019-01-26