@Divya: nope, your Q requires equal count too whereas this doesn't.

On Thu, Jun 3, 2010 at 1:06 PM, divya jain <sweetdivya....@gmail.com> wrote:

> same question as wat i asked in partioning of array such that the diff is
> min.
>
>
> On 31 May 2010 22:07, Senthilnathan Maadasamy <
> senthilnathan.maadas...@gmail.com> wrote:
>
>> Same question with interesting answers in stackoverflow :
>> http://stackoverflow.com/questions/890171/algorithm-to-divide-a-list-of-numbers-into-2-equal-sum-lists
>>
>>
>>
>> On Mon, May 31, 2010 at 5:54 PM, Anurag Sharma <anuragvic...@gmail.com>wrote:
>>
>>> Well, in that case, then just forget the "balancing the number of
>>> elements in the 2 teams" portion of my solution above :)
>>>
>>>
>>> Anurag Sharma
>>> http://anuragsharma-sun.blogspot.com/
>>>
>>>
>>> On Mon, May 31, 2010 at 10:38 AM, Nik_nitdgp 
>>> <nikhil.bhoja...@gmail.com>wrote:
>>>
>>>> This problem is like 2 processor job scheduling problem ,We may get an
>>>> optimal solution for different instances using different algorithm
>>>> apart from brute force.Whereas Brute force covers all possible subsets
>>>> but may take years to complete if N is large.
>>>>
>>>> above algo fails in the following example.
>>>>
>>>> Eg. 2 2 2 3 3
>>>>
>>>> above algo gives:
>>>> T1: 2 2 3 =7
>>>> T2: 2 3 =5
>>>>
>>>> But closest distribution is
>>>> T1=2 2 2=6
>>>> T2 3 3=6
>>>>
>>>> On May 31, 9:30 am, W Karas <wka...@yahoo.com> wrote:
>>>> > Is this the same problem as:
>>>> >
>>>> >
>>>> http://groups.google.com/group/algogeeks/browse_frm/thread/26c31cc253.
>>>> ..
>>>> >
>>>> > ?
>>>> >
>>>> > Or can the teams have different numbers of players?
>>>> >
>>>> > On May 30, 2:28 pm, Veer Sharma <thisisv...@rediffmail.com> wrote:
>>>> >
>>>> > > Hi Friends,
>>>> >
>>>> > > This is my first post to this forum. A "Hi" to all of you and here
>>>> is
>>>> > > my first problem...
>>>> >
>>>> > > Giiven int n, the total number of players and their skill-point.
>>>> > > Distribute the players on 2 evenly balanced teams.
>>>> >
>>>> > > Lets see who gives the best solution (least space complexity / least
>>>> > > time complexity or both...)
>>>>
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