Forming Distinctive Array Combos in Golang

The problem

You might be given an array of arrays and your job will likely be to return the variety of distinctive arrays that may be shaped by choosing precisely one ingredient from every subarray.

For instance: remedy([[1,2],[4],[5,6]]) = 4, as a result of it leads to solely 4 possibilites. They’re [1,4,5],[1,4,6],[2,4,5],[2,4,6].

Just be sure you don’t depend duplicates; for instance remedy([[1,2],[4,4],[5,6,6]]) = 4, for the reason that additional outcomes are simply duplicates.

The answer in Golang

Choice 1:

bundle resolution
func Resolve(information [][]int) int {
  qtd := 1
  for _, sss := vary information {
    mp := make(map[int]bool)
    for _, e := vary sss {
      mp[e] = true
    }
    qtd *= len(mp)
  }
  return qtd
}

Choice 2:

bundle resolution
func Resolve(information [][]int) int {
  res := 1
  for _, d := vary information{
    cnt := 0
    dup := make(map[int]int)
    for _, n := vary d {
      if dup[n] == 0 {
        cnt++
        dup[n] = 1
      }
    }
    res *= cnt
  }
  return res
}

Choice 3:

bundle resolution
func Resolve(information [][]int) int {
  units := make([]map[int]bool, len(information))
  for i := 0; i < len(information); i++ {
    units[i] = make(map[int]bool, len(information[i]))
    for _, v := vary information[i] {
      units[i][v] = true
    }
  }
  end result := 1
  for _, s := vary units {
    end result *= len(s)
  }
  return end result
}

Take a look at circumstances to validate our resolution

bundle solution_test
import (
  "math/rand"
  "time"
  . "github.com/onsi/ginkgo"
  . "github.com/onsi/gomega"
)
func init() {
  rand.Seed(time.Now().UTC().UnixNano())
}
var _ = Describe("Pattern Exams", func() {
  It("ought to work with pattern assessments", func() {
    Anticipate(Resolve([][]int{{1, 2}, {4}, {5, 6}})).To(Equal(4))
    Anticipate(Resolve([][]int{{1, 2}, {4, 4}, {5, 6, 6}})).To(Equal(4))
    Anticipate(Resolve([][]int{{1, 2}, {3, 4}, {5, 6}})).To(Equal(8))
    Anticipate(Resolve([][]int{{1, 2, 3}, {3, 4, 6, 6, 7}, {8, 9, 10, 12, 5, 6}})).To(Equal(72))
  })
})