## Wednesday, February 3, 2016

### Longest Zig-Zag Subsequence - GeeksforGeeks

ZigZag - TopCoder
Longest Zig-Zag Subsequence - GeeksforGeeks
The longest Zig-Zag subsequence problem is to find length of the longest subsequence of given sequence such that all elements of this are alternating.
If a sequence {x1, x2, .. xn} is alternating sequence then its element satisfy one of the following relation :
  x1 < x2 > x3 < x4 > x5 < …. xn or     x2 > x2 < x3 > x4 < x5 > …. xn
Input: arr[] = {10, 22, 9, 33, 49, 50, 31, 60}
Output: 6
The subsequences {10, 22, 9, 33, 31, 60} or
{10, 22, 9, 49, 31, 60} or {10, 22, 9, 50, 31, 60}
are longest Zig-Zag of length 6.
This problem is an extension of longest increasing subsequence problem, but requires more thinking for finding optimal substructure property in this.
We will solve this problem by dynamic Programming method, Let A is given array of length n of integers. We define a 2D array Z[n][2] such that Z[i][0] contains longest Zig-Zag subsequence ending at index i and last element is greater than its previous element and Z[i][1] contains longest Zig-Zag subsequence ending at index i and last element is smaller than its previous element, then we have following recurrence relation between them,
Z[i][0] = Length of the longest Zig-Zag subsequence
ending at index i and last element is greater
than its previous element
Z[i][1] = Length of the longest Zig-Zag subsequence
ending at index i and last element is smaller
than its previous element

Recursive Formulation:
Z[i][0] = max (Z[i][0], Z[j][1] + 1);
for all j < i and A[j] < A[i]
Z[i][1] = max (Z[i][1], Z[j][0] + 1);
for all j < i and A[j] > A[i]

The first recurrence relation is based on the fact that, If we are at position i and this element has to bigger than its previous element then for this sequence (upto i) to be bigger we will try to choose an element j ( < i) such that A[j] < A[i] i.e. A[j] can become A[i]’s previous element and Z[j][1] + 1 is bigger than Z[i][0] then we will update Z[i][0].
Remember we have chosen Z[j][1] + 1 not Z[j][0] + 1 to satisfy alternate property because in Z[j][0] last element is bigger than its previous one and A[i] is greater than A[j] which will break the alternating property if we update. So above fact derives first recurrence relation, similar argument can be made for second recurrence relation also.
// Function to return longest Zig-Zag subsequence length
int zzis(int arr[], int n)
{
    /*Z[i][0] = Length of the longest Zig-Zag subsequence
          ending at index i and last element is greater
          than its previous element
     Z[i][1] = Length of the longest Zig-Zag subsequence
          ending at index i and last element is smaller
          than its previous element   */
    int Z[n][2];
    /* Initialize all values from 1  */
    for (int i = 0; i < n; i++)
        Z[i][0] = Z[i][1] = 1;
    int res = 1; // Initialize result
    /* Compute values in bottom up manner */
    for (int i = 1; i < n; i++)
    {
        // Consider all elements as previous of arr[i]
        for (int j = 0; j < i; j++)
        {
            // If arr[i] is greater, then check with Z[j][1]
            if (arr[j] < arr[i] && Z[i][0] < Z[j][1] + 1)
                Z[i][0] = Z[j][1] + 1;
            // If arr[i] is smaller, then check with Z[j][0]
            if( arr[j] > arr[i] && Z[i][1] < Z[j][0] + 1)
                Z[i][1] = Z[j][0] + 1;
        }
        /* Pick maximum of both values at index i  */
        if (res < max(Z[i][0], Z[i][1]))
            res = max(Z[i][0], Z[i][1]);
    }
    return res;
}

http://www.cnblogs.com/lautsie/p/3250929.html

f1[i]表示 最后一个数是A[i]并且A[i]和前一个数的差值为正的最长子序列的长度。
f2[i]表示最后一个数是A[i]并且A[i]和前一个数的差值为负的最长子序列的长度。

    public int longestZigZag(int[] sequence) {
        if (sequence.length == 0) return 0;

        int[] up = new int[sequence.length];
        int[] down = new int[sequence.length];
        up[0] = 1;
        down[0] = 1;
        int ret = 0;
        for (int i = 1 ; i < sequence.length; i++) {
            int max_up = 0;
            int max_down = 0;
            for (int j = 0; j < i; j++) {
                if (sequence[j] > sequence[i]) {
                    if (down[j] > max_down) max_down = down[j];
                }
                else if (sequence[j] < sequence[i]) {
                    if (up[j] > max_up) max_up = up[j];
                }
            }
            up[i] = max_down + 1;
            down[i] = max_up + 1;

            if (up[i] > ret) ret = up[i];
            if (down[i] > ret) ret = down[i];
        }

        return ret;
    }
http://rafal.io/posts/topcoder-zigzag.html
Let A an array of length of n of integers. Let Z(i,0) be the longest zig-zag subsequence ending at index i with a positive difference (the second last element of it is strictly less than the last element), and let Z(i,1) be the longest zig-zag subsequence ending at index i with a negative difference. Then:

Z(i,0)Z(i,1)Z(0,0)Z(0,1)=minj<i,A[j]<A[i](Z(j,1)+1)=minj<i,A[j]>A[i](Z(j,0)+1)=1=1
public static int longestZigZag(int[] A){
int n = A.length;
int[][] Z = new int[n][2];
for(int i = 0; i < Z.length; i++){
Z[i] = new int[2];
}
Z[0][0] = 1;
Z[0][1] = 1;

int best = 1;

for(int i = 1; i < n; i++){
for(int j = i-1; j>= 0; j--){
if(A[j] < A[i]) Z[i][0] = Math.max(Z[j][1]+1,Z[i][0]);
if(A[j] > A[i]) Z[i][1] = Math.max(Z[j][0]+1, Z[i][1]);
}
best = Math.max(best, Math.max(Z[i][0],Z[i][1]));
}
return best;
}
https://github.com/detel/Algorithms/blob/master/DP/LongestZigZagSubsequence.java