How do I find the first two consecutive elements in my array of numbers?

Question

Using Ruby 2.4, I have an array of unique, ordered numbers, for example

[1, 7, 8, 12, 14, 15]

How do I find the first two elements whose differ

Answer 1

Here are three more ways.

#1

def first_adjacent_pair(arr)
  (arr.size-2).times { |i| return arr[i, 2] if arr[i+1] == arr[i].next }
  nil
end

first_adjacent_pair [1, 7, 8, 12, 14, 15] #=> [7,8]
first_adjacent_pair [1, 7, 5, 12, 14, 16] #=> nil

#2

def first_adjacent_pair(arr)
  enum = arr.to_enum # or arr.each
  loop do
    curr = enum.next
    nxt = enum.peek
    return [curr, nxt] if nxt == curr.next
  end
  nil
end

enum.peek raises a StopIteration exception when the enumerator enum has generated its last element with the preceding enum.next. The exception is handled by Kernel#loop by breaking out of the loop, after which nil is returned. See also Object#to_enum, Enumerator#next and Enumerator#peek.

#3

def first_adjacent_pair(arr)
  a = [nil, arr.first] 
  arr.each do |n|
    a.rotate!
    a[1] = n
    return a if a[1] == a[0] + 1
  end
  nil
end

See Array#rotate!.