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

One way to do this:

a.each_with_index { |e, i| break [ e, a[i.next] ] if a[i.next] == e.next } 
#=> [7, 8]

Unlike chunk or each_cons this doesn't create an array of arrays. It also breaks as soon as a pair is found.

---

Benchmarks

require 'fruity'

arr = ((1...1000)).to_a.reverse + [1,2]

def first_adjacent_pair(arr)
  idx = arr.each_index.drop(1).find { |i| (arr[i-1]-arr[i]).abs == 1 }
  idx ? arr[idx-1, 2] : nil
end

def first_adjacent_pair2(arr)
  enum = arr.to_enum
  loop do
    curr = enum.next
    nxt = enum.peek
    return [curr, nxt] if (curr-nxt).abs == 1
  end
  nil
end

compare do
  iceツ  { ar = arr.dup; ar.each_with_index { |e, i| break [ e, ar[i.next] ] if ar[i.next] == e.next }  }
  cary   { ar = arr.dup; first_adjacent_pair(ar) }
  cary2  { ar = arr.dup; first_adjacent_pair2(ar) }  
  seb    { ar = arr.dup; ar.each_cons(2).find{|a,b| b-a == 1} }
end

#Running each test 64 times. Test will take about 1 second.
#cary2 is faster than cary by 3x ± 0.1
#cary is faster than iceツ by 3x ± 0.1 (results differ: [999, 998] vs [1, 2])
#iceツ is faster than seb by 30.000000000000004% ± 10.0%

Answer 2

You could use each_cons and find to get the first element from the array of pairs where the second element less the first one is equal to 1:

p [1, 7, 8, 12, 14, 15].each_cons(2).find { |a, b| b - a == 1 }
# => [7, 8]

Answer 3

Here's an alternate method provided for educational purposes:

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

arr.each_cons(2).map {|v|v.reduce(:-)}.index(-1)

Answer 4

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!.

Answer 5

Simple example

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

   X.each_with_index do |item, index|
    if index < X.count - 1
     if (X[index+1]-X[index] == 1) 
      puts item
     end
   end
  end