@above...can you please enlighten me about the second term in the dp
expression
And are you sure its O(n) ?

On Fri, Jan 28, 2011 at 7:08 PM, sankalp srivastava <
[email protected]> wrote:

> This can be done in O(n) very easily , similar to Longest increasing
> subsequence
>
> Solution :-
>
> dp[l]= maximum length of the zigzag sequence upto the length l
>
> //At any position , the particular number in the array can either
> extend the zigzag sequence containing the last element or it can start
> one of it's own . So the recurrance becomes
>
> dp[i]= max(dp[j]) ,and diff[A[p[j]]]^(A[i]-A[p[j]]&31)!=0 , j<i
>
> find out the maximum in this array , it will get you the answer .
>
> PS:- You can also check out the Topcoder tutorials .
>
> On Jan 27, 7:41 pm, bittu <[email protected]> wrote:
> > well I found it  as it  Can be Done in O(n). but with additional space
> > O(n)
> > here program is written in Java
> >
> > public class ZigZag
> > {
> >
> >  public int longestZigZag(int[] sequence)
> >   {
> >   if (sequence.length==1) return 1;
> >   if (sequence.length==2) return 2;
> >   int[] diff = new int[sequence.length-1];
> >
> >   for (int i=1;i<sequence.length;i++)
> >  {
> >    diff[i-1]=sequence[i]-sequence[i-1];
> >   }
> >
> >   //90% Program is done here it self. by looking at the sign if
> > alternative number in auxiliary array we can count longest  zigzag
> > array
> >
> >   int sign=sign(diff[0]);
> >   int count=0;
> >   if (sign!=0)
> >    count=1;
> >
> >    for (int i=1;i<diff.length;i++)
> >   {
> >    int nextsign=sign(diff[i]);
> >    if (sign*nextsign==-1){
> >     sign=nextsign;
> >     count++;
> >    }
> >   }
> >   return count+1;
> >  }
> >
> >  public int sign(int a)
> >  {
> >   if (a==0) return 0;
> >   return a/Math.abs(a);
> >  }
> >
> >  public static void main(String[] args)
> >   {
> >   ZigZag z = new ZigZag();
> >   System.out.println(z.longestZigZag(new int[] { 1, 7, 4, 9, 2, 5 }));
> >   System.out.println(z.longestZigZag(new int[] { 1, 17, 5, 10, 13, 15,
> > 10, 5, 16, 8 }));
> >
> >    }
> >
> > }
> >
> > Try for Inplace
> >
> > Thanks & Regards
> > Shashank Mani ""The best way to escape from a problem is to solve it."
>
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>


-- 
S.Nishaanth,
Computer Science and engineering,
IIT Madras.

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