Ruby problem ⑦

This is the final episode.


Given the three integers a b c, b or c has a difference of 1 from a And True if the difference between the values of b and c is 2 or more. Other than that, False Let's create a method that outputs.

Output example:

close_far(1, 2, 10) → True close_far(1, 2, 3) → False close_far(4, 1, 3) → True


Use the abs method to convert the return value to an integer.

abs You can get the absolute value by executing the "abs" method on the target number. In other words, if it is a positive number, it will remain as it is, but if it is a negative number, you can get the numerical value that is made into a positive number by taking the sign.

The actual usage is as follows.

num = 5.abs
#=> 5

num = (-5).abs
#=> 5

##### Model answer
def close_far(a,b,c)
  x = (a-b).abs
  y = (a-c).abs
  z = (b-c).abs

  if x == 1 && z >= 2
    puts "True"
  elsif y == 1 && z >= 2
    puts "True"
    puts "False"

There was no explanation this time, but it was understandable by looking at the answer. The point is to substitute the difference such as a-b. I couldn't think of it. I wanted to get to the correct answer ...

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