From: "Cédric Finance" <pticedric@...>
Date: 2007-11-08T00:44:22+09:00
Subject: Re: a problem related string(250 score)

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This should solve the problem.

class DiskClusters
  def self.minimumFragmentation(str,int)
  groups = str.split(/X+/).map { |e| e.size }.sort
  res = 0
  while(val = groups.pop) do
    int -= val
    res += 1
    return res if int <= 0
  end
  -1
  end
end

Cedric

On Nov 7, 2007 4:12 PM, Johnson Wang <99100@163.com> wrote:

> How to solve this problem in Ruby????
> ==============================================
> You are given a String disk representing the clusters on a disk. An 'X'
> represents a used cluster, and a '.' represents an available cluster.
> You are also given an int size representing the size, in clusters, of a
> file waiting to be written to disk. A file can only be stored in
> clusters not already being used.
> Return the minimum number of groups of consecutive clusters needed to
> store the file on the disk. (The disk does not wrap around at the end.)
> Return -1 if the disk does not have enough space available to store the
> file.
> Definition
>
> Class:
> DiskClusters
> Method:
> minimumFragmentation
> Parameters:
> String, int
> Returns:
> int
> Method signature:
> int minimumFragmentation(String disk, int size)
> (be sure your method is public)
>
>
> Constraints
> -
> disk will contain between 1 and 50 characters, inclusive.
> -
> Each character of disk will be 'X' or '.'.
> -
> size will be between 1 and 50, inclusive.
> Examples
> 0)
>
> "."
> 2
> Returns: -1
> We can't fit the file on the disk.
> 1)
>
> ".XXXXXXXX.XXXXXX.XX.X.X."
> 6
> Returns: 6
> There is only ever one cluster together, so all six clusters are
> separated.
> 2)
>
> "XX..XX....X.XX........X...X.XX...XXXX..XX...XXXXX."
> 12
> Returns: 2
> We fit eight clusters together, and four clusters together.
> 3)
>
> ".X.XXXX.......XX....X.....X............XX.X.....X."
> 20
> Returns: 3
>
> 4)
> "....X...X..X"
> 11
> Returns: -1
> --
> Posted via http://www.ruby-forum.com/.
>
>

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