ruby - Given integers how do I find asc and desc sequences of three? -

i have integers i.e. 9, 5, 4, 3, 1, 6, 7, 8. want return index sequence of 3 descending or ascending integers exists. in example above indices 1 , 5. ruby code this?

def seq   array = [9,5,4,3,1,6,7,8]   array.each_with_index |val, index|      if (val < (array[index + 1]).val < (array[index + 1]).val)       puts "#{index}"      # skip 2 indexes      end end 

i think logic behind solution correct, syntax pretty far off valid ruby.

here pair of pretty verbose solutions (hopefully) obvious:

numbers = [9, 6, 5, 4, 3, 1, 6, 7, 8]  # find non-overlapping sets = 0 until > numbers.length - 2   a, b, c = numbers[i..i + 2]   if (a - b == b - c) && (a - b).abs == 1     puts "#{i} (#{a},#{b},#{c})"     # skip next 2 indexes     += 3   else     += 1   end end  # find overlapping sets (same solution, don't skip indexes) (0...numbers.length - 2).each |i|   a, b, c = numbers[i..i + 2]   if (a - b == b - c) && (a - b).abs == 1     puts "#{i} (#{a},#{b},#{c})"   end end