# Question

Formatted question description: https://leetcode.ca/all/376.html

```
376 Wiggle Subsequence
A sequence of numbers is called a wiggle 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 wiggle sequence.
For example,
[1,7,4,9,2,5] is a wiggle 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 wiggle sequences, the first because its first two differences are positive
and the second because its last difference is zero.
Given a sequence of integers, return the length of the longest subsequence that is a wiggle sequence.
A subsequence is obtained by deleting some number of elements (eventually, also zero) from the original sequence,
leaving the remaining elements in their original order.
Example 1:
Input: [1,7,4,9,2,5]
Output: 6
Explanation: The entire sequence is a wiggle sequence.
Example 2:
Input: [1,17,5,10,13,15,10,5,16,8]
Output: 7
Explanation: There are several subsequences that achieve this length. One is [1,17,10,13,10,16,8].
Example 3:
Input: [1,2,3,4,5,6,7,8,9]
Output: 2
Follow up:
Can you do it in O(n) time?
@tag-dp
```

# Algorithm

### DP

We maintain two dp arrays p and q,

`p[i]`

indicates the maximum length of the swing subsequence with a positive first difference when the position i is reached`q[i]`

indicates the maximum length of the wobble sub-sequence with a negative first difference when reaching the i position

We traverse the array from i=1, and then for each traversed number, traverse to this number from the beginning position, then compare nums[i] and nums[j], and update the corresponding positions respectively.

### Greedy

Maintain two variables p and q, and then traverse the array

- If the current number is greater than the previous number, then
`p=q+1`

- If the current number is smaller than the previous number, then
`q=p+1`

Finally, compare the larger value of p and q with n, and choose the smaller one.

# Code

Java