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?

Code

DP

Time Complexity: $O(n)$, Space Complexity: $O(n)$

class Solution {
public:
    int wiggleMaxLength(vector<int>& nums) {
        int n = nums.size();
        if(n == 0) return 0;
 
        int up[n], down[n];
        memset(up, 0, sizeof(up));
        memset(down, 0, sizeof(down));
 
        up[0] = 1;
        down[0] = 1;
 
        for(int i = 1; i < nums.size(); i++) {
            if((nums[i] - nums[i - 1]) > 0) {
                up[i] = down[i - 1] + 1;
                down[i] = down[i - 1];
            } else if ((nums[i] - nums[i - 1]) < 0) {
                up[i] = up[i - 1];
                down[i] = up[i - 1] + 1;
            } else {
                up[i] = up[i - 1];
                down[i] = down[i - 1];
            }
        }
 
        return max(up[n-1], down[n-1]);
       
    }
};
 
// up[i]: maxLength ends in upward slope up to nums[i]
// end[i]: maxLength ends in downward slope up to nums[i]

DP - Optimized

Time Complexity: $O(n)$, Space Complexity: $O(1)$

class Solution {
public:
    int wiggleMaxLength(vector<int>& nums) {
        int n = nums.size();
        if(n == 0) return 0;
 
        int up = 1;
        int down = 1;
 
        for(int i = 1; i < nums.size(); i++) {
            if((nums[i] - nums[i - 1]) > 0) {
                up = down + 1;
            } else if ((nums[i] - nums[i - 1]) < 0) {
                down = up + 1;
            }
        }
 
        return max(up, down);
       
    }
};
 

Source

• Wiggle Subsequence - LeetCode