# 376. Wiggle Subsequence

### Problem statement

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

```plaintext
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

```plaintext
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

```plaintext
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?

### Solution: Counting the local extrema of `nums`

First, if you pick all local extrema (minima and maxima) of `nums` to form a subsequence `e`, then it is wiggle. Let us call it an **extrema** subsequence.

#### Example 2

For `nums = [1,17,5,10,13,15,10,5,16,8]`, the local extrema are `[1,17,5,15,5,16,8]`. It is wiggle and called **extrema** subsequence.

Note that if `nums.length = n` then `nums[0]` and `nums[n - 1]` are always the first and the last extremum.

Second, given any two successive local extrema `a` and `b`, you cannot have any wiggle subsequence between them. Because the elements between them are either monotonic increasing or decreasing.

That proves the extrema subsequence is the longest wiggle one.

#### Code

```cpp
#include <iostream>
#include <vector>
using namespace std;
int wiggleMaxLength(vector<int>& nums) {
    // nums[0] is always the first extremum
    // start to find the second extremum
    int i = 1;
    while (i < nums.size() && nums[i] == nums[i - 1]) {
        i++;
    }
    if (i == nums.size()) {
        // all nums[i] are equal
        return 1;
    }
    int sign = nums[i] > nums[i - 1] ? 1 : -1;
    int count = 2;
    i++;
    while (i < nums.size()) {
        if ((nums[i] - nums[i - 1]) * sign < 0) {
            // nums[i] is an extremum
            count++;
            sign = -sign;
        }
        i++;
    }
    return count;
}
int main() {
    vector<int> nums{1,7,4,9,2,5};
    cout << wiggleMaxLength(nums) << endl;
    nums = {1,17,5,10,13,15,10,5,16,8};
    cout << wiggleMaxLength(nums) << endl;
    nums = {1,2,3,4,5,6,7,8,9};
    cout << wiggleMaxLength(nums) << endl;
}
```

```plaintext
Output:
6
7
2
```

#### Complexity

* Runtime: `O(n)`, where `n = nums.length`.
    
* Extra space: `O(1)`.
    

### References

* [https://leetcode.com/problems/wiggle-subsequence/](https://leetcode.com/problems/wiggle-subsequence/)
    

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