Determining whether one array contains the contents of another array in ruby

Question

In ruby, how do I test that one array not only has the elements of another array, but contain them in that particular order?

correct_combination = [1, 2, 3, 4, 5         


        

Answer 1

I would suggestion a for loop that compares each one

@out_of_order_elements = []

for i in 0.. @array_size do
  unless submission_array[i] == @correct_combination[i]
    @out_of_order_ids.push(@submission_array[i])
  end
end 

Answer 2

I would like to consider continuous sequence of elements from other array in container array and present this:-- code is inspired from the Sawa's code

class Array
  def contain? other
   arr = self & other
   (arr.eql? other ) && ((self.index(arr.last) - self.index(arr.first)).eql?(other.size - 1))
  end
end

Result:--

correct_combination = [1, 2, 3, 4, 5]
[1, 5, 8, 2, 3, 4, 5].contain?(correct_combination) # => false
[8, 10, 1, 2, 3, 4, 5, 9].contain?(correct_combination) # => true
[1, 8, 2, 3, 4, 5].contain?(correct_combination) # => false

Answer 3

This is what I came up with

a = [1, 2, 3, 4, 5]
b = [2, 3, 5]
c = [3, 9]

irb(main):037:0* (a + b).sort.uniq == a.sort.uniq
=> true
irb(main):038:0> (a + c).sort.uniq == a.sort.uniq
=> false

Answer 4

This is the best I could come up with. All the return calls are a bit ugly, but it should be quicker than doing a string comparison if it's large arrays.

class Array
  def same?(o)
    if self.size == o.size
      (0..self.size).each {|i| return false if self[i] != o[i] }
    else
      return false
    end

    return true
  end
end

a = [1,2,3,4,5]
b = [1, 5, 8, 2, 3, 4, 5]
c = [1, 2, 6, 4, 5]

puts a.same?(a.reverse) # => false
puts a.same?(a) # => true
puts a.same?(b) # => false
puts a.same?(c) # => false

Answer 5

You can use each_cons method:

arr = [1, 2, 3, 4, 5]
[1, 5, 8, 2, 3, 4, 5].each_cons(arr.size).include? arr

In this case it will work for any elements.