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Cover image for Oversized Pancake Flipper - Google Code Jam

Oversized Pancake Flipper - Google Code Jam

greenkey profile image Lorenzo Mele ・3 min read

Google Code Jam

If you're a programmer and you don't know Google Code Jam, you should definitely go and check that. It's basically a coding competition, I like to call it "the programming olympics".

It's about 4 or 5 years I'm participating, I love it because it challenges my coding skills, and I'm not talking about "making a for loop".

This year I want to share my solutions of the problems, let's start with the first and the easiest one.

The problem

Here the complete problem text.

You have a row of pancakes and a strange flipper, you have to find the right combination to make them show their happy side.

At first I thought it could be easy: just flip the unhappy pancakes and see what happens.

"This is a competition, the solution can't be so simple!" - my inner voice said. So I proceeded to the other problem to see if they were simpler, after those I came back.

The solution

Facing the problem again, I had no other ideas but the first one. I decided to give it a try.

You can find the code here, but let me explain some part of it.

To ease the following steps, I get the first part of the input line.split()[0] and I convert the pancake row in a list of binary items.

# ...
pancake_row = [p == '+' for p in line.split()[0]]
# ...

The list comprehension says something like "for every character in the string, create an item with True value if the character equals +, False otherwise"
List comprehension are very powerful, here is the code without it:

pancake_row = list()
for p in line.split()[0]:
    pancake_row.append(p == '+')

Then i go through the list looking for a pancake to flip, now that every item is a boolean, I can check them with a simple if:

# ...
pan_size = int(line.split()[1])
# ...
while i < (len(pancake_row) - pan_size + 1):
    if not pancake_row[i]:
# ...

Now I put the flipper starting from the first unhappy pancake and then flip.

pancakes: ---+-++-
flipper:  ***
flipped:  ++++-++-

To make it in the code, I just have to not the items: not False == True:

# ...
while i < (len(pancake_row) - pan_size + 1):
    if not pancake_row[i]:
        for j in range(i, i + pan_size):
            pancake_row[j] = not pancake_row[j]
# ...

I think I could have done it using slicing:

# ...
while i < (len(pancake_row) - pan_size + 1):
    if not pancake_row[i]:
        j = i + pan_size
        pancake_row[i:j] = [not p for p in pancake_row[i:j]]
# ...

Going on with the example, the pancake row evolution should be the following:

---+-++-
***
++++-++-
    ***
+++++---
     ***
++++++++

With this example we're lucky, because at the end all the pancakes are happy. We can check it using a special Python function (you don't have to reinvent the wheel):

# ...
all(pancake_row)
# ...

all()

Return True if all elements of the iterable are true

If the pancakes are not all on the happy side, we can argue that it's impossible: the unhappy pancake should be at the end of the row, there's no way flip them without flipping the previous pancakes.

What I learnt

My intuition was right... I was right! (not my inner voice).

So next time I could give it a try sooner.

Posted on by:

greenkey profile

Lorenzo Mele

@greenkey

Life is short. A bio is too short to describe even a part of it.

Discussion

markdown guide
 

Wow! Nice solution. I have been thinking about it for a while and I still have two open quiestions:

  • Does it work in any case?
  • Does it find the optimal solution (minimum flips)?

The first one seems to be true (at least the cases tested) but I can't see why. I am not so sure about the second one.
Can you give me any clue about them?

Due to my background I would go for a graph search based solution. The nodes of the graph are the state of the row of pancakes and the neighbors of a node are those states reachable by a flip action.
It is not the fastest one, but it is exhaustive and the solution is optimal.

 
  • Does it work in any case?

I cannot show you the mathematical explanation but I have some hints.
Let's take the example of the post but reversed:

-++-+---
***
+---+---
 ***
+++++---
     ***
++++++++

These are exactly the same flips made above.

I think the key point here is that you should never touch happy pancakes, unless forced to do so because of a nearest unhappy one.

 

This is not fair !!
I was trying to solve it with other ways.
solution is so simple that its raising Questions..?

does its a complete solution ?

  • theoretically I can see its not optimal in finding of minimum flips
  • theoretically with complex and long input(where cursor should move backward as well) its incomplete solution !
 

Huh, I can't believe the solution is so simple in the end!
I tried to solve it using XOR and discrete mathematics :(
Thanks for the post!

BTW you got a typo ("at the and")

 

Yes, that was my thought too!

(thanks for the typo ;) )

 

How many rounds do I need to pass if i want to have a job in google🤔

 

I don't know... I think you should be notable (anything that means) :)