//code sketch .... based on greedy approach.
jumps(int hop,int n)
{
if(hop > n)
{
return;
}
if(hop==n)
{
//path found
}
for(i=hop ; i<n && i< hop+arr[hop] ; i++)
{
jumps( (i+arr[i])-1 , n);
}
}
little help required to find out path in minimum hop....please do modify
code as required.
thanks
On Sat, Jan 28, 2012 at 12:48 AM, Don <dondod...@gmail.com> wrote:
> At first I thought that I needed a special case to avoid zeros.
> However, if you can move past a zero to a non-zero, that is always a
> preferred move, and if not, a move to a location before the zero which
> allows you to move past the zero is also better. If no such move
> exists, there is no way to get to the end.
> Don
>
> On Jan 27, 12:23 pm, sravanreddy001 <sravanreddy...@gmail.com> wrote:
> > @Don:
> >
> > The solution looks good...
> > I can see that the greedy choice property is holding.. and its optimal
> > too...
> >
> > max (j+a[J]) maximizing is leading us to the farthest possible position,
> >
> > but.. in the beginning.. i thought.. this will have probs with 0's
> > but.. couldn't come up an example, for which ur approach fail and there's
> > soultion for it.
>
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