Problem Description

The aim of this problem is to find the length of the longest wiggle subsequence in a given integer array, nums. A wiggle sequence is defined as one where consecutive numbers alternate between increasing and decreasing. That means the difference between successive numbers should switch between positive and negative. Single-element sequences and two-element sequences with distinct values are also considered wiggle sequences.

For instance, [1, 7, 4, 9, 2, 5] is a wiggle sequence because the differences between consecutive numbers (6, -3, 5, -7, 3) switch signs from positive to negative and vice versa. On the other hand, [1, 4, 7, 2, 5] is not a wiggle sequence because it starts with two positive differences. Similarly, [1, 7, 4, 5, 5] isn't either because it ends with a difference of zero.

A subsequence is derived by removing some elements from the original sequence while maintaining the relative order of the remaining elements. The problem asks to return the maximum length of such a longest wiggle subsequence.

Intuition

The intuition behind solving this problem lies in dynamic programming. The solution leverages two counters, up and down, to keep track of the two scenarios while traversing the array:

• up: This counter is incremented when the current element is greater than the preceding one, indicating an "up" wiggle. If this occurs, the length of the longest wiggle sequence considering the current number as an "up" number is the current value of down plus one. In other words, it's extending a sequence that was last going "down" with an "up" number.

• down: Conversely, this counter is incremented when the current element is less than the preceding one, signifying a "down" wiggle. In this situation, the length of the longest wiggle sequence considering the current number as a "down" number is derived from the current length of up plus one. That is, we're adding a "down" number to a subsequence that was previously "up".

An important point to note is that for both up and down, we use the max() function to ensure that we are always recording the maximum length of wiggle subsequence found so far – effectively making this a dynamic programming solution where the optimal solution of the current step is built upon the previous steps. At each step, we make a decision based on the current and previous elements to update these counters.

At the end of the traversal, the length of the longest wiggle subsequence will be the maximum value between the up and down counters because we are interested in the longest possible sequence, regardless of whether it ends on an "up" wiggle or a "down" wiggle.

Learn more about Greedy and Dynamic Programming patterns.

Solution Approach

The implementation of the solution is grounded in the principles of dynamic programming, which is an optimization technique that solves complex problems by breaking them down into simpler subproblems.

In this specific problem, the dynamic programming pattern used involves two one-dimensional arrays, up and down, where each entry up[i] or down[i] would normally represent the length of the longest wiggle subsequence up to the i-th index of nums that ends with an increasing or decreasing wiggle, respectively. The reference solution compresses these arrays into two simple integer variables because it cleverly leverages the fact the current state only depends on the previous state.

Here's how the algorithm progresses:

1. Initialize two variables up and down to 1, since a single element is trivially a wiggle subsequence.
2. Loop through the elements of the array starting from the second element because we're comparing each element with its previous one.
3. For each element at index i, compare it with the element at index i - 1.
   • If nums[i] is greater than nums[i-1], it means we have a potential "up" wiggle. We then update the up counter as follows:

     up = max(up, down + 1)

     This is under the notion that the current "up" wiggle can extend a sequence that was previously "down", hence down + 1.
   • If nums[i] is less than nums[i - 1], it signals a potential "down" wiggle and down is updated similarly:

     down = max(down, up + 1)

     Here, the current "down" wiggle can extend a sequence that was previously "up".
4. As the for loop continues, up and down are updated dynamically with respect to the sign alternation between consecutive elements.
5. After completing the traversal, the largest of up or down is returned as the maximum length of the longest wiggle subsequence found.

The reason behind this simple and elegant implementation is that for every position in the array nums[i], the up and down states captures the length of the longest wiggle subsequence that ends with a "down" or "up" wiggle up to that position, respectively. The two possible cases of a wiggle (up or down) are covered by these two variables. There's no need to maintain an additional array because the maximum lengths of "up" and "down" wiggle subsequences at any position only depend on the immediate previous one.

Therefore, this algorithm is efficient in terms of both time complexity, which is O(n) since it only requires a single pass through the array, and space complexity, which is O(1) because it maintains only a constant number of variables.

Example Walkthrough

Let's consider the following input array: nums = [1, 17, 5, 10, 13, 15, 10, 5, 16, 8]. We want to find the length of the longest wiggle subsequence.

Following the solution approach:

1. Initialize two variables up and down to 1. Without loss of generality, we can think of every single number as a trivial wiggle subsequence.

2. Start loop from index 1, since we can only start determining a wiggle by comparing at least two numbers:

   • At i = 1 (nums[1] = 17, nums[0] = 1):
     • As nums[1] > nums[0], we have an "up" wiggle. Update up:

       up = max(up, down + 1) = max(1, 1 + 1) = 2

   • At i = 2 (nums[2] = 5, nums[1] = 17):
     • As nums[2] < nums[1], we have a "down" wiggle. Update down:

       down = max(down, up + 1) = max(1, 2 + 1) = 3

   • At i = 3 (nums[3] = 10, nums[2] = 5):
     • As nums[3] > nums[2], we have an "up" wiggle. Update up:

       up = max(up, down + 1) = max(2, 3 + 1) = 4

3. Continue this process for each element in the array. The updates will proceed as follows:

   • At i = 4:
     • nums[4] = 13, nums[3] = 10, so up remains 4 (13 continues the "up" sequence from 10).
   • At i = 5:
     • nums[5] = 15, nums[4] = 13, so up remains 4 (15 continues the "up" sequence from 13).
   • At i = 6:
     • nums[6] = 10, nums[5] = 15, we update down to 5 (as 10 is down from 15).
   • At i = 7:
     • nums[7] = 5, nums[6] = 10, down remains 5 (5 continues the "down" sequence from 10).
   • At i = 8:
     • nums[8] = 16, nums[7] = 5, we update up to 6 (as 16 is up from 5).
   • At i = 9:
     • nums[9] = 8, nums[8] = 16, we update down to 7 (as 8 is down from 16).

4. By the end of the loop, the up and down variables have been updated accordingly. The maximum length of the longest wiggle subsequence is the larger one of up and down, which in this case is down = 7.

This example illustrates how the approach keeps updating the two counters that represent the length of the longest possible wiggle subsequences based on whether the current element goes "up" or "down" relative to the previous element in the subsequence. With this method, we can effectively solve the problem in linear time with constant extra space.

Solution Implementation

Time and Space Complexity

The given Python code snippet implements the solution for finding the length of the longest wiggle subsequence in an array.

Time Complexity

To analyze the time complexity, let's look at the operations performed by the code. The code iterates through the nums list once using a single loop that starts from 1 up to len(nums) (non-inclusive). Inside this loop, there are only constant-time operations: comparisons, arithmetic operations, and max function calls on integers.

As the loop runs for n-1 iterations (where n is the length of nums) and each operation within the loop takes constant time, the overall time complexity of the code is O(n).

Space Complexity

Regarding space complexity, the code uses a fixed number of integer variables (up and down). No matter the size of the input, the space used by these variables does not increase. There are no additional data structures like lists or sets that grow with the size of the input.

Therefore, the space complexity of the code is O(1), which means it's constant space complexity.

Learn more about how to find time and space complexity quickly using problem constraints.