@divya :  descending order sorting works. BRILLIANT !!

On 5/17/10, divya jain <[email protected]> wrote:
> my algo on the array 1 200 500 2000
> sort the array therefore we have now 2000 500 200 1
> 1st array will have largest element
> A= {2000}
> and B={500}
> sumA=2000
> sumB=500
> now abs((2000+200)-500)>abs((2000)-(500+200))
> so we ll put 200 in array B. now since B has n/2 elements rest of the
> element goes to array which is 1.
> so the ans is
> A={2000,1}
> b={500,200}
>
>
> On 15 May 2010 19:10, Rohit Saraf <[email protected]> wrote:
>
>> so what will ur algo give for array 1,200,500,2000
>>
>> On 5/15/10, divya jain <[email protected]> wrote:
>> > my approach:
>> > 1. sort the array
>> > 2. take a variable diff. intitialize it to 0.
>> > 3. take the 1st element from array nd place it in array A and 2nd
>> > element
>> in
>> > array B. stoe in diff sum(A)-sum(B).
>> > 4. now place the next element in array A or B according to the condition
>> if
>> >        if sum(A+element)-sum(B)> sum(a)-sum(B+element). store the
>> > element
>> in
>> > B  otherwise in A. also while storing the element in ny array maintain
>> the
>> > count of element in that aaray. if any time the count reaches n/2 where
>> > n
>> is
>> > the no. of elements in  the given aaray. then store rest element in the
>> > other array.
>> > 5. repeat step 5 until both array A n B get n/2 elements..
>> >
>> > hope my approach is clear and correct.
>> > comments are welcomed.....
>> >
>> > On 15 May 2010 08:47, Rohit Saraf <[email protected]> wrote:
>> >
>> >> Choosing a greedy strategy for this would be difficult.
>> >>
>> >> For a simple dp you can
>> >> maintain A[i,total,present] using a recurrence
>> >>
>> >> i is the present index of array
>> >> total is the number of elements reqd in first partition.
>> >> present is the no of elements already there in first partition.
>> >>
>> >> the array stores difference between sums. GET the minimum of all these
>> >> and backtrack.
>> >>
>> >>
>> >> On 5/15/10, Amir hossein Shahriari <[email protected]>
>> >> wrote:
>> >> > @karas: your solution is greedy and its wrong e.g. for
>> >> > {1,2,3,4,5,100}
>> >> your
>> >> > diff would be 95 but the best result is 91
>> >> >
>> >> > i think we can solve this problem by dynamic programming but not a
>> >> > simple
>> >> > one! since the size of the two subsets must be equal.
>> >> > so it's DP solution has at least 3 dimensions: tow dimensions
>> >> representing
>> >> > the number of elements in each subset and another for the difference
>> >> between
>> >> > their sums
>> >> >
>> >> > On Fri, May 14, 2010 at 10:11 PM, W Karas <[email protected]> wrote:
>> >> >
>> >> >> On May 14, 4:51 am, divya <[email protected]> wrote:
>> >> >> > Algorithm to partition set of numbers into two s.t. diff bw their
>> >> >> > sum is min and they hav equal num of elements
>> >> >>
>> >> >> void part(const int a[], int n_a, int g1[], int g2[])
>> >> >>  {
>> >> >>    int i, j, k;
>> >> >>
>> >> >>    /* diff = sum(g1) - sum(g2) */
>> >> >>    int diff;
>> >> >>
>> >> >>    sort(a, n_a);
>> >> >>
>> >> >>    diff = 0;
>> >> >>    for (i = 0, j = 1, k = 0; j < n_a; ++k, i += 2, j += 2)
>> >> >>      {
>> >> >>        if ((a[i] > a[j]) == (diff > 0))
>> >> >>          {
>> >> >>            g1[k] = a[j];
>> >> >>            g2[k] = a[i];
>> >> >>          }
>> >> >>        else
>> >> >>          {
>> >> >>            g1[k] = a[i];
>> >> >>            g2[k] = a[j];
>> >> >>          }
>> >> >>        diff += g1[k] - g2[k];
>> >> >>       }
>> >> >>  }
>> >> >>
>> >> >> --
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>> >> --------------------------------------------------
>> >> Rohit Saraf
>> >> Second Year Undergraduate,
>> >> Dept. of Computer Science and Engineering
>> >> IIT Bombay
>> >> http://www.cse.iitb.ac.in/~rohitfeb14<http://www.cse.iitb.ac.in/%7Erohitfeb14>
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>> --------------------------------------------------
>> Rohit Saraf
>> Second Year Undergraduate,
>> Dept. of Computer Science and Engineering
>> IIT Bombay
>> http://www.cse.iitb.ac.in/~rohitfeb14<http://www.cse.iitb.ac.in/%7Erohitfeb14>
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Rohit Saraf
Second Year Undergraduate,
Dept. of Computer Science and Engineering
IIT Bombay
http://www.cse.iitb.ac.in/~rohitfeb14

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