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Asked: 2026-06-02T21:57:19+00:00

I’ve been trying to think since HOURS about this TopCoder problem and couldn’t come

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I’ve been trying to think since HOURS about this TopCoder problem and couldn’t come with a perfectly working solution and found the one given below that is insanely beautifully used!

I’m trying to figure how this solution works for the given probem? And how could I have originally thought of it? After reading the solution I figured it’s a variant of Huffman coding but that’s as far as I could get. I’m really enthralled and would like to know what line of thought could lead to this solution..

Here’s the problem:
http://community.topcoder.com/stat?c=problem_statement&pm=11860&rd=15091

Fox Ciel has lots of homework to do. The homework consists of some
mutually independent tasks. Different tasks may take different amounts
of time to complete. You are given a int[] workCost. For each i, the
i-th task takes workCost[i] seconds to complete. She would like to
attend a party and meet her friends, thus she wants to finish all
tasks as quickly as possible.

The main problem is that all foxes, including Ciel, really hate doing
homework. Each fox is only willing to do one of the tasks. Luckily,
Doraemon, a robotic cat from the 22nd century, gave Fox Ciel a split
hammer: a magic gadget which can split any fox into two foxes.

You are given an int splitCost. Using the split hammer on a fox is
instantaneous. Once a hammer is used on a fox, the fox starts to
split. After splitCost seconds she will turn into two foxes — the
original fox and another completely new fox. While a fox is splitting,
it is not allowed to use the hammer on her again.

The work on a task cannot be interrupted: once a fox starts working on
a task, she must finish it. It is not allowed for multiple foxes to
cooperate on the same task. A fox cannot work on a task while she is
being split using the hammer. It is possible to split the same fox
multiple times. It is possible to split a fox both before and after
she solves one of the tasks.

Compute and return the smallest amount of time in which the foxes can
solve all the tasks.

And here’s the solution I found at this link

import java.util.*; 

public class FoxAndDoraemon { 
  public int minTime(int[] workCost, int splitCost) { 
    PriorityQueue<Integer> pq = new PriorityQueue<Integer>(); 

    for(int i : workCost) pq.offer(i); 

    while(pq.size()>=2) { 
      int i = pq.poll(); 
      int j = pq.poll(); 
      pq.offer(Math.max(i, j) + splitCost); 
    } 
    return pq.poll(); 

  } 
}
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1 Answer

  1. Editorial Team

    First of all you do realize the reasoning behind `max(i, j) + splitCost’. Don’t you? Basically, if you have one fox, you split it into two and perform each task independently. Let us call this process “merging”.

    Now assume we have three jobs a,b and c such that a>b>c. You can either do merge(merge(a,b),c) or merge(merge(a,c),b) or merge(merge(b,c),a). Do the math and you can prove that merge(merge(b,c),a) is least among these three.

    You can now use induction to prove that this solution is valid for any number of jobs (not just 3).

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