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Facebook Hacker Cup Qualification Round

12 Jan

So. I have a final later today and I’m too bored of studying so I decided to write up a post on the latest Facebook Hacker Cup competition problems. As you may know, this past weekend, Facebook had the Qualification Round of their Hacker Cup. Basically, it’s a programming contest very similar to that of Google Code Jam that I mentioned a while back and follows a pretty regular format.

In this round, we only had to answer one question out of three correctly to qualify. I did qualify so I know I got at least one right. Strangely, they only tell you the results after the competition is over, which is both weird and annoying. Oh and they messed up sending out the emails to the people who qualified. I made an incredibly dumb error in that I accidentally downloaded the input file before I was done writing my code and so the 6-minute time limit expired before I could submit my result. Fortunately, Facebook lifted this six-minute restriction due to bugs in their system so my (eventual correct) solution was actually accepted. The problem description and my solution is listed below:

Double Squares

I can’t actually access the original question right now so I’m paraphrasing here. Effectively, we are given an input file with the first line being N, the number of lines that follow. Each of those lines contain an integer between 0 and 2147395600, inclusive. (Not 100% sure if that’s the right number exactly but its within that range). The solution should contain one integer per line of input containing the number of ways that the number can be written (order doesn’t matter) as a sum of two squares.

Solution

There was a very nice little hint in light gray at the top of the problem page that effectively said:

too hard for brute force, switch to dp

Nice. Gladly, I saw that and immediately switched to a smarter algorithm (I wasn’t sure how hard these problems were and I wasn’t going to waste time writing a sophisticated algorithm if it was unnecessary – although my solution is pretty straightforward for this problem).

First, I used a calculator to compute the square root of that number. It was in the order of about 46,000, which is still small enough that pre-computing a data structure of that size is not out of the question. Further, since that number fits comfortably within the range of double-precision numbers, there was nothing special I needed to do. I started off by generating an ArrayList<Double> of each of the numbers from 0 to 46,342 squared. This actually executed faster than I thought it would and so I continued along this idea. (Note: For use in the DP algorithm I will describe, a HashSet would be a faster data structure although since I couldn’t perform a binary search on it, I decided to use an ArrayList. Since this is performed only once per input line, I didn’t stress too much about it.) Next, having performed the binary search of the input number on my giant list, I computed and stored the index of the greatest value less than or equal to the number itself.

With this index, I created a second array which contained numbers only less than or equal to it. The logic here is that since all our numbers are non-negative (indeed, the squares of all real numbers are non-negative), each of the addends must be less than or equal to the number (remember, 0 is on the list too). Now having generated this sub-array, I traverse it and subtract each element of it from the input number itself. This is effectively subtracting one of the squares from the number. What is the reason for this? To check if the remaining quantity is a perfect square and if so, to note that the resulting pair is one that we are seeking. I used some quick and dirty counting scheme of adding a half for each time I saw this (since we will be hitting upon commutative pairs) and adding a whole number for the number itself. Then we simply return the count that we have accumulated.

Very simple algorithm and runs pretty fast. We do a constant amount of pre-processing so total time for that \mathcal{O}(1). Now if our number is K, then binary searching for it runs in time complexity \mathcal{O}(1) time (remember, the size of the big array is fixed). Creating the sub-array and traversing it and subtracting it from the original number runs in \mathcal{O}(\sqrt{K}) time and traversing the list to search for numbers in the original list runs in \mathcal{O}(\sqrt{K} \cdot 1) = \mathcal{O}(\sqrt{K}) time (remember again, the big array has a fixed size!). Thus, our total running time complexity for each query (input number) is also \mathcal{O}(\sqrt{K}). One optimization would be to not search the big list for a match for every number in the sub-array but rather just until \lfloor \frac{\sqrt{K}}{2} \rfloor indices and then multiplying the count by two. In addition, if I had used a HashSet for the big array instead, we would enjoy amortized constant time searching which would give us a time complexity of \mathcal{O}(\sqrt{K}), which is the same big-Oh time complexity but much faster if you write it in tilde notation.

import java.util.*;
import java.io.*;

public class Main
{
    int SIZE = 46342;
    ArrayList<Double> squares;

    public Main()
    {
        squares = new ArrayList()<Double>;

        for (int i = 0; i < SIZE; i++)
            squares.add(Math.pow(i, 2));

        solve();
    }

    private void solve()
    {
        Scanner in = new Scanner(System.in);

        int N = in.nextInt();

        for (int i = 0; i < N; i++)
            output(in.nextDouble());
    }

    private void output(double i)
    {
        if (i <= 1)         
        {            
           System.out.println("1");             
           return;         
        }
        
        double num = 0.0;
        int index = Collections.binarySearch(squares, i);
        if (index > 0)
            num++;

        index = (index >= 0) ? index: Math.abs(index) - 1;

        ArrayList subList = new ArrayList();

        for (int j = 1; j < index; j++)
            subList.add(i - squares.get(j));

        for (int j = 0; j < index - 1; j++)
             if (Collections.binarySearch(squares, subList.get(j)) >= 0)
                num += 0.5;

        System.out.println((int)num);
    }

    public static void main(String args[])
    {
        Main app = new Main();
    }
}

On my computer, running Ubuntu 32-bit and OpenJDK with Core 2 Duo processors at 2.53GHz, the total time taken to run this program on the input file given was 0.264s.

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2 Comments

Posted by on January 12, 2011 in Uncategorized

 

Tags: , , , , ,

2 responses to “Facebook Hacker Cup Qualification Round

  1. ankit

    January 16, 2011 at 11:40 AM

    Look at this great blog explaining the solutions of qualification round of Hacker Cup.
    Double Squares: http://itbhu.ac.in/codefest/blog/?p=159
    Peg Game: http://itbhu.ac.in/codefest/blog/?p=172
    Studious Student : http://itbhu.ac.in/codefest/blog/?p=180
    Hope you will love this.

     

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