Check if array is an ordered subset

Question

I want to find out if an array is an ordered subset of another array:

• [1,2] is an ordered subset of [1,2,3]
• [1,3] is an ordered subset of [1,2,3]
• [2,1] is not an ordered subset of [1,2,3]

I've found some solutions to this, but every solution ignores the order. Every method I've seen so far ignores the order of the arrays:

[1,2,3] - [2,1] #=> [3]
[1,2,3] & [2,1] #=> [1,2]
[1,2,3].to_set.superset?([2,1].to_set) #=> true

Update: Based on the discussion below, I've updated my question.

Answer 1

b == a & b

That checks whether b is contained in a in the same order.

In other words: In general you have B⊆A ⇔ B=A∩B. And Ruby's Array#& preserves the order (of the left operand).

Answer 2

a = [1,2,3]
b = [2,1]

p a.each_cons(b.size).any?{|slice| slice == b} # => false

Answer 3

Given two arrays, arr and sub, this is a way to determine if there exists an array of strictly increasing indices, indices, such that arr.values_at(*indices) == sub.

def ordered?(arr, sub)
  sub.each do |c|
    i = arr.index(c)
    return false if i.nil?
    arr = arr[i+1..-1]
  end
  true
end

ordered?([1,2,3], [1,2])           #=> true
ordered?([1,2,3], [2,3])           #=> true
ordered?([1,2,3], [1,3])           #=> true
ordered?([1,2,3], [3,1])           #=> false
ordered?([1,2,5,2,4,3,4], [2,2,3]) #=> true

Note that @StefanPochmann suggested a more compact way of writing this in a comment below.