# 376. Wiggle Subsequence

## Description

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?

## Solutions

### 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.

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

• class Solution {
public:
int wiggleMaxLength(vector<int>& nums) {
int up = 1, down = 1;
for (int i = 1; i < nums.size(); ++i) {
if (nums[i] > nums[i - 1]) {
up = max(up, down + 1);
} else if (nums[i] < nums[i - 1]) {
down = max(down, up + 1);
}
}
return max(up, down);
}
};

• class Solution:
def wiggleMaxLength(self, nums: List[int]) -> int:
up = down = 1
for i in range(1, len(nums)):
if nums[i] > nums[i - 1]:
up = max(up, down + 1)
elif nums[i] < nums[i - 1]:
down = max(down, up + 1)
return max(up, down)

############

class Solution(object):
def wiggleMaxLength(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if not nums:
return 0
up = down = 1
for i in range(1, len(nums)):
if nums[i] > nums[i - 1]:
up = down + 1
elif nums[i] < nums[i - 1]:
down = up + 1
return max(up, down)


• func wiggleMaxLength(nums []int) int {
up, down := 1, 1
for i := 1; i < len(nums); i++ {
if nums[i] > nums[i-1] {
up = max(up, down+1)
} else if nums[i] < nums[i-1] {
down = max(down, up+1)
}
}
return max(up, down)
}

• function wiggleMaxLength(nums: number[]): number {
let up = 1,
down = 1;
for (let i = 1; i < nums.length; ++i) {
let prev = nums[i - 1],
cur = nums[i];
if (cur > prev) {
up = Math.max(up, down + 1);
} else if (cur < prev) {
down = Math.max(down, up + 1);
}
}
return Math.max(up, down);
}