This means , You can make a sum = j , with or without using the item i ,
while calculating P[i][j].
So you can have another counter Count2 which will have the count for such
items. So you will calculate P as discussed before
but You will add 1 in Count2[i][j] whenever you find that case. add one in
count[i][j] in any of P = 1 case.
in the end you'll search for the max value of j (closest to S/2) for all P
values which has this property on value of i .
1. i = 50
2. for all i> 50
i-count2[i][j] <= 50
I think this will do. Check it out.
On Thu, Dec 30, 2010 at 12:41 PM, Ankur Khurana <[email protected]>wrote:
> vishal , what will we do to count when both p[i-1][j] and
> p[i-1][j-a[i]] is true .
>
> On Thu, Dec 30, 2010 at 12:36 PM, Ankur Khurana
> <[email protected]> wrote:
> > Thanks everybody for wonderful support and special thanks to Vishal
> > raja. . But i was bit apprehensive about your last solution . . i will
> > test it :) and let you know as well . Thanks . . . .
> >
> >
> > On Thu, Dec 30, 2010 at 11:52 AM, vishal raja <[email protected]>
> wrote:
> >> But the same solution I've given above can give you the solution for
> this
> >> problem .
> >> In the formed table of P[i][j] , you can take another variable attached
> to
> >> it as count[i][j] for how many items we have selected yet.
> >> So you gotta find , the max. value of j which has count = 50.
> >> count[i][j] = count[i-1][j] if P(i-1,j) ==1
> >> count[i][j] = count[i-1][j-a[i]] if P(i-1,j-a[i]) ==1
> >> else count[i][j] = 0
> >>
> >>
> >>
> >>
> >> On Thu, Dec 30, 2010 at 11:42 AM, vishal raja <[email protected]>
> >> wrote:
> >>>
> >>> yeah, My bad.
> >>> Missed that.
> >>>
> >>> On Wed, Dec 29, 2010 at 10:52 PM, Wladimir Tavares <
> [email protected]>
> >>> wrote:
> >>>>
> >>>> Sum up all the number and divide by 2
> >>>>
> >>>> Using the algorithm subset problem to find a number close to median
> >>>>
> >>>>
> >>>> Wladimir Araujo Tavares
> >>>> Federal University of Ceará
> >>>>
> >>>>
> >>>>
> >>>>
> >>>>
> >>>>
> >>>> On Wed, Dec 29, 2010 at 2:07 PM, Ankur Khurana <
> [email protected]>
> >>>> wrote:
> >>>>>
> >>>>> How will you divide and array of approx 100 elements into two sub
> sets
> >>>>> of 50 each such that the difference between both the subsets is the
> >>>>> minimum possible one . .
> >>>>>
> >>>>> Thanks in advance .
> >>>>> Ankur
> >>>>>
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