TopCoder Dynamic Programming Longest Zig Zag Sequence Problem

Dynamic Programming - Longest Zig Zag Sequence

Problem 

ZigZag - 2003 TCCC Semifinals 3

A sequence of numbers is called a zig-zag sequence if the differences between successive numbers strictly alternate between positive and negative. The first difference (if one exists) may be either positive or negative. A sequence with fewer than two elements is trivially a zig-zag sequence.

For example, 1,7,4,9,2,5 is a zig-zag sequence because the differences (6,-3,5,-7,3) are alternately positive and negative. In contrast, 1,4,7,2,5 and 1,7,4,5,5 are not zig-zag sequences, the first because its first two differences are positive and the second because its last difference is zero.

Given a sequence of integers, sequence, return the length of the longest subsequence of sequence that is a zig-zag sequence. A subsequence is obtained by deleting some number of elements (possibly zero) from the original sequence, leaving the remaining elements in their original order.

Constraints
- sequence contains between 1 and 50 elements, inclusive.
- Each element of sequence is between 1 and 1000, inclusive.

Examples
   
{ 1, 7, 4, 9, 2, 5 }
Returns: 6
The entire sequence is a zig-zag sequence.
   
{ 1, 17, 5, 10, 13, 15, 10, 5, 16, 8 }
Returns: 7
There are several subsequences that achieve this length. One is 1,17,10,13,10,16,8.

Solution

int findLongZigZag(int *a, int size) {

  if(size<=2) {

        printf("%d\n", size);

        return;

  }

  // to track alternate positive and negative sequence.

  int *b = (int *)malloc(sizeof(int)*size);

  // to track length of longest sequence till ith index. 

  int *s = (int *)malloc(sizeof(int)*size));

  int i,j;

  s[0]=1;

  b[0]=a[0];

  b[1]=a[1]-a[0];

  s[1]=2;

  for(i=2;i<size;i++) {

        s[i]=2;

        b[i]=0;

  }

  for(i=2;i<size;i++) {

     for(j=1;j<i;j++) {

        if(b[j]<0) {

           if(a[i] > a[j] && s[i]<=s[j]+1) {

                s[i]=s[j]+1;

                b[i]=a[i]-a[j];

           }

        } else {

          if(a[i] < a[j] && s[i]<=s[j]+1) {

                s[i]=s[j]+1;

                b[i]=a[i]-a[j];

          }

        }

     }

     if(b[i]==0) {

        b[i]=a[i]-a[i-1];

     }

  }

  return s[size-1];

}