@Nitin Garg well if negatives are included there would be infinite
number of solutions
right?

and we can start we start with dividing the sum by combinations.
Atleast one number must be greater than sum/combination..
Am not sure it this is same as that subset manipulation...
pls post the algo for that recursion method!

On Oct 27, 12:21 am, Nitin Garg <[email protected]> wrote:
> Are we talking about only positive integers here?
>
> On Wed, Oct 26, 2011 at 11:33 PM, Vaibhav Mittal
> <[email protected]>wrote:
>
>
>
>
>
>
>
>
>
> > +1 Prem
> > @ligerdave : I knew about the recursion method..but can u throw some light
> > on the pointer based method..(with a small example maybe)..
> > Specifically I wanted to know the implementation part and the running time
> > of the algorithm.
>
> > On Wed, Oct 26, 2011 at 8:33 PM, ligerdave <[email protected]> wrote:
>
> >> @meng You already have the pattern figured out. each time subtract 1
> >> from the lowest digit and add to higher digit(only once), until the
> >> lowest digit equals to closest higher digit. the selection of which
> >> number to start could be figured out with given parameters sum and
> >> combination
>
> >> @Prem, no recursion needed here. it make it more complex than
> >> necessary. one loop with a pointer should be able to resolve this
>
> >> On Oct 24, 6:28 pm, Meng Yan <[email protected]> wrote:
> >> > Hi, my question is
>
> >> > given sum=N and combination constraint=M (the number of elements), how
> >> to
> >> > find all possible combinations of integers?
>
> >> > For example, given sum=6, combination=3; how to get the result as
> >> following:
> >> > 1+1+4;
> >> > 1+2+3;
> >> > 2+2+2;
>
> >> > We don't care about order of the elements, which means 1+1+4 and 1+4+1
> >> are
> >> > considered as same combination.
>
> >> > Thanks a lot!
>
> >> > Meng
>
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> --
> Nitin Garg
>
> "Personality can open doors, but only Character can keep them open"

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