Wiggle Subsequence#

Problem statement#

[1]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?

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#

#include <iostream>
#include <vector>
using namespace std;
int wiggleMaxLength(const 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;
}

Output:
6
7
2

Complexity#

  • Runtime: O(n), where n = nums.length.

  • Extra space: O(1).

Conclusion#

The problem of finding the length of the longest wiggle subsequence can be efficiently solved using a greedy approach. The solution iterates through the input array, identifying alternating extremums (peaks and valleys) to form the wiggle subsequence.

By keeping track of the current trend direction (increasing or decreasing), the solution efficiently identifies extremums and increments the count accordingly. This greedy approach ensures that each extremum contributes to increasing the length of the wiggle subsequence, maximizing its overall length.