## Question

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,
`[1, 7, 4, 9, 2, 5]`

is a**wiggle sequence**because the differences`(6, -3, 5, -7, 3)`

alternate between positive and negative. - In contrast,
`[1, 4, 7, 2, 5]`

and`[1, 7, 4, 5, 5]`

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 `nums`

, return *the length of the longest **wiggle subsequence** of* `nums`

.

**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:**

`1 <= nums.length <= 1000`

`0 <= nums[i] <= 1000`

**Follow up:** Could you solve this in `O(n)`

time?

## Algorithm

- Get the diffs array helps you to think through this questions but you don't really need this array when solving the question
- So you count from start and if there is wiggle, res ++, else ignore it until next wiggle
- In my not working case, I didn't handle the up/down carefully cause the wiggle can start from down or up

- Handling the "up" and "down" separately is necessary because the problem requires the differences between successive numbers to strictly alternate between positive and negative. By keeping track of the lengths of the longest wiggle subsequences ending in an upward and downward direction separately, we ensure that we consider both possibilities at each step of the iteration.(ChatGPT provided )

## Code(DP)

class Solution { public int wiggleMaxLength(int[] nums) { if (nums == null || nums.length <= 1) return nums.length; int up = 1, down = 1; int i = 1; while (i < nums.length) { if (nums[i] < nums[i - 1]) down = up + 1; else if (nums[i] > nums[i - 1]) up = down + 1; i++; } return Math.max(up, down); } }

## Code2(Not working)

class Solution { public int wiggleMaxLength(int[] nums) { if (nums == null || nums.length == 0) { return 0; } else if (nums.length == 1) { return 1; } int[] diffs = new int[nums.length-1]; boolean noWiggle = true; for (int i = 1; i < nums.length; i++) { diffs[i - 1] = nums[i] - nums[i - 1]; if (diffs[i-1] != 0) { noWiggle = false; } } int res = 2; for (int i = 1; i < diffs.length; i++) { if (diffs[i] * diffs[i-1] < 0) { res++; } } return noWiggle ? 1 : res; } }

## Code2(Greedy)

public class Solution { public int wiggleMaxLength(int[] nums) { if (nums.length < 2) return nums.length; int prevdiff = nums[1] - nums[0]; int count = prevdiff != 0 ? 2 : 1; for (int i = 2; i < nums.length; i++) { int diff = nums[i] - nums[i - 1]; if ((diff > 0 && prevdiff <= 0) || (diff < 0 && prevdiff >= 0)) { count++; prevdiff = diff; } } return count; } }