matt lao
matt lao

Reputation: 361

What is wrong with my && operator?

I would like to know why this is outputting false when I input 1982. Is there something wrong with my && statement? I tried using !(t==r), but it didn't work; for some reason, it keeps outputting false.

  def no_repeats?(year)
   out=true
   t=0
   while t<4
     r=0
     while r<4
       if (year[t] == year[r]) && t != r
         out=false
       end
         r+=1

      end
      t+=1
   end

   out
  end

Upvotes: 0

Views: 68

Answers (1)

kobaltz
kobaltz

Reputation: 7070

You're probably complicating this a bit more than it needs to be.

2.2.0-preview1 :001 > load 'no_repeat.rb'
 => true

Testing as a string

2.2.0-preview1 :002 > no_repeats?("1981")
 => false
2.2.0-preview1 :003 > no_repeats?("1983")
 => true

Testing as an integer

2.2.0-preview1 :004 > no_repeats?(1981)
 => false
2.2.0-preview1 :005 > no_repeats?(1983)
 => true

and no_repeat.rb looks like

  def no_repeats?(year)
    digits = year.to_s.split(//)
    digits.size == digits.uniq.size
  end

Edit: Benchmarks

Using Original Post

real    0m0.598s
user    0m0.583s
sys     0m0.015s



Using .split(//)

real    0m1.322s
user    0m1.321s
sys     0m0.000s



Using .chars.to_a

real    0m0.562s
user    0m0.557s
sys     0m0.004s

So, in efforts to make this more of a complete answer, I've included my benchmarks, each using the method 400,000 times. By using split(//), you will be taking almost a 2x performance hit. By using chars.to_a instead, you'll be up with your original speeds.

def no_repeats?(year)
  digits = year.to_s.chars.to_a
  digits.size == digits.uniq.size
end

Upvotes: 2

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