@above
[20, 20, 3, 42]

regards
- Sumit Kumar Pathak
(Sumit/ Pathak/ SKP ...)
*Smile is only good contagious thing.*
*Spread it*!!!!!


On Mon, Feb 4, 2013 at 7:40 AM, navneet singh gaur <
[email protected]> wrote:

> I guess it doesn't require a DP, I might have understood your question
> wrongly but from what I have understood solution is as follows :
>
> S = sum at a particular point
> A[N] = array which contains guard's respective demands
>
> i = 0, S=0;
> while (i < N)
> {
>    if (A[i] >= S)
>    {
>       S += A[i];
>    }
> i++;
> }
>
>
> On Mon, Feb 4, 2013 at 5:54 PM, marti <[email protected]> wrote:
> > You have N guards in a line each with a demand of coins.You can skip
> paying
> > a guard only if his demand is lesser than what you have totally paid
> before
> > reaching him.Find the least number of coins you spend to cross all
> guards.
> > I think its a DP problem but cant come up with a formula.Another approach
> > would be to binary search on the answer but how do I verify if no. of
> coins
> > is a possible answer?
> >
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>
>
> --
> navneet singh gaur
>
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