=begin
5 kyu

Description:
Sum of Pairs
Given a list of integers and a single sum value, return the first two values (parse from the left please) in order of appearance that add up to form the sum.

sum_pairs([11, 3, 7, 5],         10)
#              ^--^      3 + 7 = 10
== [3, 7]

sum_pairs([4, 3, 2, 3, 4],         6)
#          ^-----^         4 + 2 = 6, indices: 0, 2 *
#             ^-----^      3 + 3 = 6, indices: 1, 3
#                ^-----^   2 + 4 = 6, indices: 2, 4
#  * entire pair is earlier, and therefore is the correct answer
== [4, 2]

sum_pairs([0, 0, -2, 3], 2)
#  there are no pairs of values that can be added to produce 2.
== None/nil/undefined (Based on the language)

sum_pairs([10, 5, 2, 3, 7, 5],         10)
#              ^-----------^   5 + 5 = 10, indices: 1, 5
#                    ^--^      3 + 7 = 10, indices: 3, 4 *
#  * entire pair is earlier, and therefore is the correct answer
== [3, 7]

Negative numbers and duplicate numbers can and will appear.

NOTE:
There will also be lists tested of lengths upwards of 10,000,000 elements.
Be sure your code doesn't time out.
=end

require 'set'

def sum_pairs(ints, s)
  set = Set.new([])
  ints.each do |int|
    if (set.include?(s - int)) then
      return [s - int, int]
    else
      set.add(int)
    end
  end
  return nil
end

l1= [1, 4, 8, 7, 3, 15]
l2= [1, -2, 3, 0, -6, 1]
l3= [20, -13, 40]
l4= [1, 2, 3, 4, 1, 0]
l5= [10, 5, 2, 3, 7, 5]
l6= [4, -2, 3, 3, 4]
l7= [0, 2, 0]
l8= [5, 9, 13, -3]


sum_pairs(l5, 10) == [3, 7]