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 <shashank7andr...@gmail.com> 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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