This is the question I am facing at level 3 of Google Foobar:

There are two types: Mach bombs (M) and Facula bombs (F). The bombs, once released into the LAMBCHOP's inner workings, will automatically deploy to all the strategic points you've identified and destroy them at the same time.

But there's a few catches. First, the bombs self-replicate via one of two distinct processes:

Every Mach bomb retrieves a sync unit from a Facula bomb; for every Mach bomb, a Facula bomb is created;

Every Facula bomb spontaneously creates a Mach bomb.

For example, if you had 3 Mach bombs and 2 Facula bombs, they could either produce 3 Mach bombs and 5 Facula bombs, or 5 Mach bombs and 2 Facula bombs. The replication process can be changed each cycle.

Second, you need to ensure that you have exactly the right number of Mach and Facula bombs to destroy the LAMBCHOP device. Too few, and the device might survive. Too many, and you might overload the mass capacitors and create a singularity at the heart of the space station - not good!

And finally, you were only able to smuggle one of each type of bomb - one Mach, one Facula - aboard the ship when you arrived, so that's all you have to start with. (Thus it may be impossible to deploy the bombs to destroy the LAMBCHOP, but that's not going to stop you from trying!)

You need to know how many replication cycles (generations) it will take to generate the correct amount of bombs to destroy the LAMBCHOP. Write a function answer(M, F) where M and F are the number of Mach and Facula bombs needed. Return the fewest number of generations (as a string) that need to pass before you'll have the exact number of bombs necessary to destroy the LAMBCHOP, or the string "impossible" if this can't be done! M and F will be string representations of positive integers no larger than 10^50. For example, if M = "2" and F = "1", one generation would need to pass, so the answer would be "1". However, if M = "2" and F = "4", it would not be possible.

# Languages

To provide a Python solution, edit solution.py

To provide a Java solution, edit solution.java

# Test cases

Inputs:

(string) M = "2"
(string) F = "1"


Output:

(string) "1"


Inputs:

(string) M = "4"
(string) F = "7"


Output:

(string) "4"


Since it's given that the inputs are number in strings and can be as long as $10^{50}$, I am using the BigInteger class in Java. The approach I have used is giving me TLE. Can anyone tell me how I can optimize it further?

import java.math.BigInteger;

public static String answer(String M, String F) {

if(M.equals("1") && F.equals("1"))
{
return "0";
}
BigInteger ans = func(new BigInteger(M),new BigInteger(F));
if(ans.compareTo(new BigInteger("-1")) == 0)
{
return "impossible";
}
else
{
return ans.toString();
}
}
public static BigInteger func(BigInteger m,BigInteger f)
{
int check;
BigInteger steps = new BigInteger("0");
BigInteger one = new BigInteger("1");
while((check = m.compareTo(f)) != 0)
{
if(check == -1)
{
f = f.subtract(m);
}
else
{
m = m.subtract(f);
}
if(m.compareTo(one) == 0 || f.compareTo(one) == 0)
break;
}
int mComparedTof = m.compareTo(f);
int mCompareToOne = m.compareTo(one);
int fCompareToOne = f.compareTo(one);
if(mCompareToOne == 0 && fCompareToOne == 0)
{
return steps;
}
else if(mComparedTof == 0 && mCompareToOne != 0)
{
return new BigInteger("-1");
}
else if(mCompareToOne == 0)
{
return steps;
}
else
{
}
}

}

• I had some problems understanding the task at first, but I realized that this looks exactly like this old problem. Maybe you will find my answer there helpful. Or some of the other answers. Commented Jun 25, 2017 at 12:21

1. Use the modulo operator. At the moment you will likely end up doing repeated subtractions from the same number (e.g. 1.6248e42 and 1.234e4) until the other one is smaller. Instead, always change the bigger number into biggerNumber modulo smallerNumber and add biggerNumber div smallerNumber to steps (which I greatly doubt needs to be a BigInteger, a long should suffice.

2. Use BigInteger.valueOf or BigInteger.ZERO or BigInteger.ONE

According to Solution in this gist

One sure way to solve this challenge is to reverse engineer the two numbers they provide. From the rules to get from step n to step n-1 we just do F = F - M or M = M - F depending if F or M is larger than the other.

For example:

F = 7   M = 4
F > M ∴ F = 7 - 4
F = 3   M = 4
M > F ∴ M = 4 - 3
F = 3   M = 1
F > M ∴ F = 3 - 1
F = 2   M = 1
F > M ∴ F = 2 - 1
F = 1   M = 1


Well.. that is fairly easy, we have a solution already - we know that if we do not end up with 1 and 1 there is no solution.

But... It is pretty slow. The specification mentions that the inputs can be up to 10^50 in size. Now imagine if we had the inputs F = 1 and M = 10^50 we would have to do our calculation 10^50 - 1 times, and who knows how long you will be waiting for that to compute.

This problem occurs when F | M is larger by several multiples than its counterpart. So how about we see how many times F | M fits into M | F, then we increase the counter for how many times it can be divided into the other one.

For example:

F = 31   M = 4
F > M ∴ F = 31 - 4 * (31 / 4) //Rounded down of course
F = 3    M = 4
...


This solution avoids the problem with massive differences in the numbers and greatly optimizes the solution.

I think this challenge is pretty straight forward but the real challenge is making it optimized, I think there is enough in this file for you to go and take a good shot at this problem. In short the steps for this solution are:

• Find out if the current step you are on is solvable.

• If it is, find out if F or M is bigger.

• Divide the smaller one into the bigger one, round down, to find out the multiplier to increase the counter and to subtract the larger one to find the answer faster.

• Repeat steps until you have 1 and 1, frequently check if solvable using a custom function and finally print out the counter of how many steps it took.

### Some tips:

• Write a custom method to check if you can actually solve the problem, i.e. check if > 0, check F != M ect.

• Expect numbers larger than 10^32! So use BigInteger for example.

• If you are testing your solution and the tests are not running and instead you get a generic error it is most likely because the solution you have provided is unoptimized and is too slow. Test out your solution with big numbers.

• If you want some good numbers to test edge case, provide a number N and N + 1 as long as they are greater than 1. Good luck!

You have a lot of subtract operations in the loop. They are going in cycle and can take a lot of time for 10⁵ numbers. So you can pre-calculate them.

Just replace

m = m.subtract(f);


with

steps = steps.add(m.subtract(m.mod(f)).divide(f).subtract(BigInteger.ONE));
m = m.subtract(f.multiply(m.subtract(m.mod(f)).divide(f)));