Daily Challenge #301 - Array Combos

In this challenge, you will be given an array of arrays and your task will be to return the number of unique arrays that can be formed by picking exactly one element from each subarray.

Examples

For example: solve([[1,2],[4],[5,6]]) = 4, because it results in only 4 possibilities. They are [1,4,5],[1,4,6],[2,4,5],[2,4,6].

Make sure that you don't count duplicates; for example solve([[1,2],[4,4],[5,6,6]]) = 4, but the extra outcomes are just duplicates.

Tests

solve([[1,2],[4,4],[5,6,6]])
solve([[1,2],[3,4],[5,6]])
solve([[1,2,3],[3,4,6,6,7],[8,9,10,12,5,6]])

Good luck!

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This challenge comes from KenKamau on CodeWars. Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

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Comments

['']

JS

const solve = (arrays) => arrays.map(array => new Set(array).size).reduce((a,b) => a * b)

[Michael Baas]

A solution in Dyalog APL:
(In APL-Terminology, these "arrays" are called "vectors" and I'm using the default notation for 'em. It would also be possible to pass a JSON-Array, but that'd 5 more characters!)

solve← {×/≢¨∪¨⍵} 

using it:

      solve(1 2)(4)(5 6)
4
      solve(1 2)(4)(5 5 6)
4
      solve(1 2)(4 4)(5 6 6)
4
      solve(1 2)(3 4)(5 6)
8
      solve(1 2 3)(3 4 6 6 7)(8 9 10 12 5 6)
72

Try it online!

[Valts Liepiņš]

My solutions in Haskell.

The first solution - inefficient(O(n) = n^3), yet very elegant looking:

import Data.List (nub)

solve :: Eq a => [[a]] -> Int
solve = product . map (length . nub)

Slightly better solution (O(n) = n^2 * log n) using containers library:

import qualified Data.Set as Set

solve :: Ord a => [[a]] -> Int
solve = product . map (Set.size . Set.fromList)

And best solution (O(n) = n^2), but accepts only integers as input:

import qualified Data.IntSet as Set

solve :: [[Int]] -> Int
solve = product . map (Set.size . Set.fromList)

[peter279k]

Here is the simple solution with Python:

def solve(arr):
    result = 1
    for sets in arr:
        sets = list(set(sets))
        result *= len(sets) / 1

    return result