yes we would use 2 dimensional array to store the values inbetween but the
main question here is that is the reccurence realation correct??
On Mon, May 9, 2011 at 11:31 AM, anshu anshumishra6...@gmail.com wrote:
@gaurav both things are same, matrix is a simple and efficient way to
do problem
@gaurav
f(i, j) = {(total number we need to make sum j including ith number in
array), if its not possible than -1};
f(i, j) = f(i-1, j-ar[i]) + 1 -- if (i-1, j-ar[i]) != -1;
answer will be the largest j for which f(i, j) will be exactly equals to
n/2;
there is something more in this but when
@anshu
then shall we use the foll reccurence realation??
1i=n
S=(s+1)/2
f(i,S)= min( f(i-1, S), f(i-1, S- ar[i]) )
On Mon, May 9, 2011 at 10:42 AM, anshu anshumishra6...@gmail.com wrote:
Use KnapSack DP. suppose the sum of element = s and number of elements
= n make a 2-dimesnsional array
@cegprakash: it doesnt ensure that both the arrays will be having same
number of elements
On Sat, May 7, 2011 at 10:10 PM, cegprakash cegprak...@gmail.com wrote:
simple..
sum all the n elements..
a= floor(sum/2)
b=sum-a
output is a b
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You received this message because you are
sorry, i misunderstood the problem statement
On Sun, May 8, 2011 at 10:53 AM, Aakash Johari aakashj@gmail.comwrote:
@cegprakash: it doesnt ensure that both the arrays will be having same
number of elements
On Sat, May 7, 2011 at 10:10 PM, cegprakash cegprak...@gmail.com wrote:
