# LC 376. Wiggle Subsequence

### [Problem link](https://leetcode.com/problems/wiggle-subsequence/)

###### tags: `leedcode` `python` `c++` `medium` `Greedy` `DP`

A **wiggle sequence** is a sequence where 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 one element and a sequence with two non-equal elements are trivially wiggle sequences.

- For example, <code>[1, 7, 4, 9, 2, 5]</code> is a **wiggle sequence** because the differences <code>(6, -3, 5, -7, 3)</code> alternate between positive and negative.
- In contrast, <code>[1, 4, 7, 2, 5]</code> and <code>[1, 7, 4, 5, 5]</code> are not wiggle sequences. The first is not because its first two differences are positive, and the second is not because its last difference is zero.

A **subsequence** is obtained by deleting some elements (possibly zero) from the original sequence, leaving the remaining elements in their original order.

Given an integer array <code>nums</code>, return the length of the longest **wiggle subsequence** of <code>nums</code>.

**Example 1:**

Input: nums = [1,7,4,9,2,5]
Output: 6
Explanation: The entire sequence is a wiggle sequence with differences (6, -3, 5, -7, 3).

**Example 2:**

Input: nums = [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] with differences (16, -7, 3, -3, 6, -8).

**Example 3:**

Input: nums = [1,2,3,4,5,6,7,8,9]
Output: 2

**Constraints:**

- <code>1 <= nums.length <= 1000</code>
- <code>0 <= nums[i] <= 1000</code>

**Follow up:** Could you solve this in <code>O(n)</code> time?

## Solution 1 - Greedy
#### Python

class Solution:
    def wiggleMaxLength(self, nums: List[int]) -> int:
        res = 1

        preDiff = 0
        curDiff = 0

        for i in range(1, len(nums)):
            curDiff = nums[i] - nums[i - 1]
            if curDiff * preDiff <= 0 and curDiff != 0:
                res += 1
                preDiff = curDiff

        return res

#### C++

        int res = 1;
        int preDiff = 0;
        int curDiff = 0;

        for (int i = 1; i < nums.size(); i++) {
            curDiff = nums[i] - nums[i - 1];
            if (curDiff * preDiff <= 0 && curDiff != 0) {
                res++;
                preDiff = curDiff;
            }
        }
        return res;

## Solution 2 - DP
#### Python

class Solution:
    def wiggleMaxLength(self, nums: List[int]) -> int:
        up = 1
        down = 1

        for i in range(1, len(nums)):
            if nums[i] > nums[i - 1]:
                up = down + 1
            if nums[i] < nums[i - 1]:
                down = up + 1
                
        return max(up, down)

#### C++

class Solution {
public:
    int wiggleMaxLength(vector<int>& nums) {
        int up = 1;
        int down = 1;

        for (int i = 1; i < nums.size(); i++) {
            if (nums[i] > nums[i - 1]) {
                up = down + 1;
            }
            if (nums[i] < nums[i - 1]) {
                down = up + 1;
            }
        }
        return max(up, down);
    }
};

>### Complexity
>| | Time Complexity | Space Complexity |
>| ----------- | --------------- | ---------------- |
>| Solution 1 | O(n) | O(1) |
>| Solution 2 | O(n) | O(1) |

## Note
x