@above : call jumps(0,n-1);
On Sat, Jan 28, 2012 at 2:00 PM, atul anand <[email protected]> wrote:
> //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 <[email protected]> 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 <[email protected]> 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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>
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